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Integration

1.8 Trigonometric Integrals

29 problems · hints, answers and solutions shown beside each one

Recall that we are using log⁡x\log x to denote the logarithm of xx with base ee. In other courses it is often denoted ln⁡x\ln x.

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Suppose you want to evaluate ∫0π/4sin⁡xcos⁡nx dx\displaystyle\int_0^{\pi/4} \sin x \cos^n x ~\dee{x} using the substitution u=cos⁡xu=\cos x. Which of the following need to be true for your substitution to work?

  1. nn must be even

  2. nn must be odd

  3. nn must be an integer

  4. nn must be positive

  5. nn can be any real number

Hint

Go ahead and try it!

Answer

(e)

Full solution

If u=cos⁡xu=\cos x, then du=−sin⁡x dx\dee{u}=-\sin x\,\dee{x}. If n≠−1n \neq -1, then

∫0π/4sin⁡xcos⁡nx dx=−∫11/2undu=[−1n+1un+1]11/2=1n+1(1−12n+1)\begin{align*}\int_0^{\pi/4} \sin x \cos^n x ~\dee{x}&= - \int_1^{1/\sqrt2} u^n \dee{u} = \left[-\frac{1}{n+1}u^{n+1}\right]_1^{1/\sqrt2}=\frac{1}{n+1}\left(1-\frac{1}{\sqrt{2}^{n+1}}\right)\end{align*}

If n=−1n=-1, then

∫0π/4sin⁡xcos⁡nx dx=−∫11/2undu=−∫11/21udu=[−log⁡∣u∣]11/2=−log⁡(12)=12log⁡2\begin{align*}\int_0^{\pi/4} \sin x \cos^n x ~\dee{x}&= - \int_1^{1/\sqrt2} u^n \dee{u} = - \int_1^{1/\sqrt2} \frac{1}{u} \dee{u} =\bigg[ -\log|u|\bigg]_1^{1/\sqrt2} \\ & =-\log\left(\frac{1}{\sqrt2}\right) = \frac{1}{2}\log 2\end{align*}

So, (e) nn can be any real number.

Q2Stage 1

Evaluate ∫sec⁡nxtan⁡xdx\displaystyle\int \sec^n x \tan x \dee{x}, where nn is a strictly positive integer.

Hint

Use the substitution u=sec⁡xu=\sec x.

Answer

1nsec⁡nx+C\dfrac{1}{n}\sec^n x +C

Full solution

We use the substitution u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u}=\sec x \tan x ~\dee{x}.

∫sec⁡nxtan⁡xdx=∫sec⁡n−1x⋅sec⁡xtan⁡x dx=∫un−1du\begin{align*}\int \sec^n x \tan x \dee{x}&=\int \sec^{n-1}x \cdot \sec x \tan x~\dee{x} = \int u^{n-1}\dee{u}\end{align*}

Since nn is positive, n−1≠−1n-1 \neq -1, so we antidifferentiate using the power rule.

=unn+C=1nsec⁡nx+C\begin{align*}&=\frac{u^n}{n}+C = \frac{1}{n}\sec^n x +C\end{align*}
Q3Stage 1

Derive the identity tan⁡2x+1=sec⁡2x\tan^2 x +1 = \sec^2 x from the easier-to-remember identity sin⁡2x+cos⁡2x=1\sin^2x+\cos^2 x =1.

Hint

Divide both sides of the second identity by cos⁡2x\cos^2 x.

Answer

We divide both sides by cos⁡2x\cos^2 x, and simplify.

sin⁡2x+cos⁡2x=1sin⁡2x+cos⁡2xcos⁡2x=1cos⁡2xsin⁡2xcos⁡2x+1=sec⁡2xtan⁡2x+1=sec⁡2x\begin{align*} \sin^2x+\cos^2 x &=1 \\\frac{\sin^2x+\cos^2 x }{\cos^2 x}&=\frac{1}{\cos^2 x} \\\frac{\sin^2x}{\cos^2 x}+1&=\sec^2 x \\\tan^2 x+1&=\sec^2 x \end{align*}
Full solution

We divide both sides by cos⁡2x\cos^2 x, and simplify.

sin⁡2x+cos⁡2x=1sin⁡2x+cos⁡2xcos⁡2x=1cos⁡2xsin⁡2xcos⁡2x+1=sec⁡2xtan⁡2x+1=sec⁡2x\begin{align*} \sin^2x+\cos^2 x &=1 \\\frac{\sin^2x+\cos^2 x }{\cos^2 x}&=\frac{1}{\cos^2 x} \\\frac{\sin^2x}{\cos^2 x}+1&=\sec^2 x \\\tan^2 x+1&=\sec^2 x \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Questions 4 through 10 deal with powers of sines and cosines. Review Section 1.8.1 in the CLP-2 text for integration strategies.

Q4Stage 2Past exam · M105 2015A

Evaluate ∫cos⁡3x dx\displaystyle\int\cos^3x\,\dee{x}.

Hint

See Example 1.8.6 in the

CLP-2 text. Note that the power of cosine is odd, and the power of sine is even (it's zero).

Answer

sin⁡x−sin⁡3x3+C\sin x-\dfrac{\sin^3 x}{3} +C

Full solution

The power of cosine is odd, and the power of sine is even (zero). Following the strategy in the text, we make the substitution u=sin⁡xu=\sin x, so that du=cos⁡x dx\dee{u}=\cos x\,\dee{x} and cos⁡2x=1−sin⁡2x=1−u2\cos^2 x = 1-\sin^2 x = 1-u^2:

∫cos⁡3x dx=∫(1−sin⁡2x)cos⁡x dx=∫(1−u2) du=u−u33+C=sin⁡x−sin⁡3x3+C\begin{align*} \int \cos^3x\,\dee{x} &=\int (1-\sin^2x)\cos x\,\dee{x} =\int (1-u^2)\,\dee{u}\\ &=u-\frac{u^3}3+C =\sin x-\frac{\sin^3 x}{3}+C \end{align*}
Q5Stage 2Past exam · 2014D

Evaluate ∫0πcos⁡2x dx\displaystyle\int_0^\pi\cos^2x\,\dee{x}.

Hint

See Example 1.8.7 in the

CLP-2 text. All you need is a helpful trig identity.

Answer

π2\dfrac{\pi}{2}

Full solution

Using the trig identity cos⁡2x=1+cos⁡(2x)2\cos^2 x=\dfrac{1+\cos(2x)}{2}, we have

∫cos⁡2xdx=12∫0π[1+cos⁡(2x)]dx=12[x+12sin⁡(2x)]0π=π2\begin{alignat*}{3} \int \cos^2 x\dee{x} &= \frac{1}{2}\int_0^\pi \big[1+\cos(2x)\big]\dee{x} &= \frac{1}{2} \Big[x+\frac{1}{2}\sin(2x)\Big]_0^\pi &=\frac{\pi}{2} \end{alignat*}
Q6Stage 2Past exam · 2016Q3

Evaluate ∫sin⁡36t cos⁡3t dt\displaystyle\int\sin^{36}t\,\cos^3t\,\dee{t}.

Hint

The power of cosine is odd, so we can reserve one cosine for du\dee{u}, and turn the rest into sines using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1.

Answer

sin⁡37t37−sin⁡39t39+C\dfrac{\sin^{37}t}{37}-\dfrac{\sin^{39}t}{39}+C

Full solution

Since the power of cosine is odd, following the strategies in the text, we make the substitution u=sin⁡tu=\sin t, so that du=cos⁡t dt\dee{u}=\cos t\,\dee{t} and cos⁡2t=1−sin⁡2t=1−u2\cos^2 t = 1-\sin^2 t = 1-u^2.

∫sin⁡36tcos⁡3t dt=∫sin⁡36t (1−sin⁡2t)cos⁡t dt=∫u36(1−u2) du=u3737−u3939+C=sin⁡37t37−sin⁡39t39+C\begin{align*} \int\sin^{36}t\cos^3t\,\dee{t} &=\int\sin^{36}t \, (1-\sin^2t)\cos t\,\dee{t} =\int u^{36}(1-u^2)\,\dee{u}\\ &=\frac{u^{37}}{37}-\frac{u^{39}}{39}+C =\frac{\sin^{37}t}{37}-\frac{\sin^{39}t}{39}+C \end{align*}
Q7Stage 2

Evaluate ∫sin⁡3xcos⁡4x dx\displaystyle\int \dfrac{\sin^3 x}{\cos^4 x} ~\dee{x}.

Hint

Since the power of sine is odd (and positive), we can reserve one sine for du\dee{u}, and turn the rest into cosines using the identity sin⁡2+cos⁡2x=1\sin^2 + \cos^2 x =1.

Answer

13cos⁡3x−1cos⁡x+C\dfrac{1}{3\cos^3 x} - \dfrac{1}{\cos x}+C

Full solution

Since the power of sine is odd (and positive), we can reserve one sine for du\dee{u}, and turn the rest into cosines using the identity sin⁡2+cos⁡2x=1\sin^2 + \cos^2 x =1. This allows us to use the substitution u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x~\dee{x}, and sin⁡2x=1−cos⁡2x=1−u2\sin^2 x = 1-\cos^2 x = 1-u^2.

∫sin⁡3xcos⁡4x dx=∫sin⁡2xcos⁡4xsin⁡x dx=∫−1−u2u4du=∫(−1u4+1u2) du=13u3−1u+C=13cos⁡3x−1cos⁡x+C\begin{align*} \int \dfrac{\sin^3 x}{\cos ^4 x} ~\dee{x}&=\int \frac{\sin^2 x}{\cos^4 x}\sin x~\dee{x} =\int -\frac{1-u^2}{u^4}\dee{u}\\ &=\int\left( -\frac{1}{u^{4}}+\frac{1}{u^{2}}\right)~\dee{u} = \frac{1}{3u^3}-\frac{1}{u}+C\\ &=\frac{1}{3\cos^3 x} - \frac{1}{\cos x}+C \end{align*}
Q8Stage 2

Evaluate ∫0π/3sin⁡4x dx\displaystyle\int_0^{\pi/3} \sin^{4}x~\dee{x}.

Hint

When we have even powers of sine and cosine both, we use the identities in the last two lines of Equation 1.8.3 in the CLP-2 text.

Answer

π8−9364\displaystyle\frac{\pi}{8} -\frac{9\sqrt3}{64}

Full solution

Both sine and cosine have even powers (four and zero, respectively), so we don't have the option of using a substitution like u=sin⁡xu=\sin x or u=cos⁡xu=\cos x. Instead, we use the identity sin⁡2θ=1−cos⁡(2θ)2\sin^2 \theta = \dfrac{1-\cos(2\theta)}{2}.

∫0π/3sin⁡4x dx=∫0π/3(sin⁡2x)2 dx=∫0π/3(1−cos⁡(2x)2)2 dx=14∫0π/3(1−2cos⁡(2x)+cos⁡2(2x)) dx=14∫0π/3(1−2cos⁡(2x)) dx+14∫0π/3cos⁡2(2x) dx\begin{align*}\int_0^{\pi/3} \sin^{4}x~\dee{x} &= \int_0^{\pi/3} \left(\sin^{2}x\right)^2~\dee{x} = \int_0^{\pi/3} \left(\frac{1-\cos(2x)}{2}\right)^2~\dee{x}\\ &=\frac{1}{4}\int_0^{\pi/3}\left(1-2\cos(2x)+\cos^2(2x)\right)~\dee{x} \\&=\frac{1}{4}\int_0^{\pi/3}\left(1-2\cos(2x)\right)~\dee{x}+\frac{1}{4}\int_0^{\pi/3}\cos^2(2x)~\dee{x}\end{align*}

We can antidifferentiate the first integral right away. For the second integral, we use the identity cos⁡2θ=1+cos⁡(2θ)2\cos^2 \theta = \dfrac{1+\cos(2\theta)}{2}, with θ=2x\theta=2x.

=14[x−sin⁡(2x)]0π/3+18∫0π/3(1+cos⁡(4x)) dx=14[π3−32]+18[x+14sin⁡(4x)]0π/3=14[π3−32]+18[π3−38]=π8−9364\begin{align*}&=\frac{1}{4}\Big[x - \sin(2x)\Big]_{0}^{\pi/3} + \frac{1}{8}\int_0^{\pi/3}(1+\cos(4x))~\dee{x}\\ &=\frac{1}{4}\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]+\frac{1}{8}\Big[x+\frac{1}{4}\sin(4x)\Big]_0^{\pi/3}\\ &=\frac{1}{4}\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]+\frac{1}{8}\left[\frac{\pi}{3}-\frac{\sqrt{3}}{8}\right]\\ &=\frac{\pi}{8} -\frac{9\sqrt3}{64}\end{align*}
Q9Stage 2

Evaluate ∫sin⁡5x dx\displaystyle\int \sin^{5}x~\dee{x}.

Hint

Since the power of sine is odd, you can use the substitution u=cos⁡xu=\cos x.

Answer

−cos⁡x+23cos⁡3x−15cos⁡5x+C-\cos x + \dfrac{2}{3}\cos^3 x - \dfrac{1}{5}\cos^5 x +C

Full solution

Since the power of sine is odd, we can reserve one sine for du\dee{u}, and change the remaining four into cosines. This sets us up to use the substitution u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x~\dee{x}.

∫sin⁡5x dx=∫sin⁡4x⋅sin⁡x dx=∫(1−cos⁡2x)2sin⁡x dx=−∫(1−u2)2 du=−∫(1−2u2+u4)du=−u+23u3−15u5+C=−cos⁡x+23cos⁡3x−15cos⁡5x+C\begin{align*} \int \sin^{5}x~\dee{x}&=\int \sin^4 x \cdot \sin x~\dee{x} = \int (1-\cos^2 x)^2 \sin x~\dee{x} \\&=-\int (1-u^2)^2~\dee{u} = -\int (1-2u^2+u^4)\dee{u}\\ &=-u+\frac{2}{3}u^3-\frac{1}{5}u^5+C\\ &=-\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x +C \end{align*}
Q10Stage 2

Evaluate ∫sin⁡1.2xcos⁡x dx\displaystyle\int \sin^{1.2}x\cos x ~\dee{x}.

Hint

Which substitution will work better: u=sin⁡xu=\sin x, or u=cos⁡xu=\cos x?

Answer

12.2sin⁡2.2x+C\dfrac{1}{2.2}\sin^{2.2}x+C

Full solution

If we use the substitution u=sin⁡xu=\sin x, then du=cos⁡x dx\dee{u}=\cos x~\dee{x}, which very conveniently shows up in the integrand.

∫sin⁡1.2xcos⁡x dx=∫u1.2du=u2.22.2+C=12.2sin⁡2.2x+C\begin{align*} \int \sin^{1.2}x\cos x ~\dee{x}&=\int u^{1.2}\dee{u} = \frac{u^{2.2}}{2.2}+C = \frac{1}{2.2}\sin^{2.2}x+C \end{align*}

Note this is exactly the strategy described in the text when the power of cosine is odd. The non-integer power of sine doesn't cause a problem.

Questions 12 through 21 deal with powers of tangents and secants. Review Section 1.8.2 in the CLP-2 text for strategies.

Q11Stage 2

Evaluate ∫tan⁡xsec⁡2xdx\displaystyle\int \tan x \sec^2 x \dee{x}.

Hint

Try a substitution.

Answer

12tan⁡2x+C\dfrac{1}{2}\tan^2 x+C, or equivalently, 12sec⁡2+C\dfrac{1}{2}\sec^2 +C

Full solution

  • Let's use the substitution u=tan⁡xu=\tan x, du=sec⁡2x dx\dee{u} = \sec^2 x~\dee{x}:

    ∫tan⁡xsec⁡2xdx=∫u du=12u2+C=12tan⁡2x+C\int \tan x \sec^2 x \dee{x} = \int u~\dee{u} = \frac{1}{2}u^2+C = \frac{1}{2}\tan^2 x +C

  • We can also use the substitution u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u} = \sec x \tan x~\dee{x}:

    ∫tan⁡xsec⁡2xdx=∫u du=12u2+C=12sec⁡2x+C\int\tan x \sec^2 x \dee{x} = \int u~\dee{u} = \frac{1}{2}u^2+C = \frac{1}{2}\sec^2 x +C

We note that because tan⁡2x\tan^2x and sec⁡2x\sec^2 x only differ by a constant, the two answers are equivalent.

Q12Stage 2Past exam · 2015A

Evaluate ∫tan⁡3xsec⁡5x dx\displaystyle\int \tan^3 x \sec^5x \,\dee{x}.

Hint

For practice, try doing this in two ways, with different substitutions.

Answer

17sec⁡7x−15sec⁡5x+C\dfrac{1}{7}\sec^7 x -\dfrac{1}{5}\sec^5 x + C

Full solution

  • Substituting u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x\,\dee{x}, sin⁡2x=1−cos⁡2x=1−u2\sin^2 x= 1-\cos^2x=1-u^2, gives

    ∫tan⁡3xsec⁡5x dx=∫sin⁡3xcos⁡8x dx=∫(1−cos⁡2x)sin⁡xcos⁡8x dx=−∫1−u2u8 du=−[u−7−7−u−5−5]+C=17sec⁡7x−15sec⁡5x+C\begin{align*} \int \tan^3 x \sec^5x \,\dee{x} &=\int\frac{\sin^3 x}{\cos^8 x}\,\dee{x} =\int\frac{(1-\cos^2 x)\sin x}{\cos^8 x}\,\dee{x} =-\int\frac{1-u^2}{u^8}\,\dee{u} \\ &=-\Big[\frac{u^{-7}}{-7}-\frac{u^{-5}}{-5}\Big]+C =\frac{1}{7}\sec^7 x -\frac{1}{5}\sec^5 x + C \end{align*}

  • Alternatively, substituting u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u}=\sec x\tan x\,\dee{x}, tan⁡2x=sec⁡2x−1=u2−1\tan^2 x= \sec^2x-1=u^2-1, gives

    ∫tan⁡3xsec⁡5x dx=∫tan⁡2xsec⁡4x (tan⁡xsec⁡x) dx=∫(u2−1)u4 du=[u77−u55]+C=17sec⁡7x−15sec⁡5x+C\begin{align*} \int \tan^3 x \sec^5x \,\dee{x} &=\int \tan^2 x \sec^4x\ (\tan x\sec x)\,\dee{x} =\int (u^2-1) u^4\,\dee{u} \\ &=\Big[\frac{u^{7}}{7}-\frac{u^{5}}{5}\Big]+C =\frac{1}{7}\sec^7 x -\frac{1}{5}\sec^5 x + C \end{align*}
Q13Stage 2Past exam · 2016Q3

Evaluate ∫sec⁡4x tan⁡46x dx\displaystyle\int\sec^4x\,\tan^{46}x\,\dee{x}.

Hint

A substitution will work. See Example 1.8.14 in the

CLP-2 text for a template for integrands with even powers of secant.

Answer

tan⁡49x49+tan⁡47x47+C\displaystyle\frac{\tan^{49}x}{49}+\frac{\tan^{47}x}{47}+C

Full solution

Use the substitution u=tan⁡xu=\tan x, so that du=sec⁡2x dx\dee{u}=\sec^2 x\,\dee{x}:

∫sec⁡4x tan⁡46x dx=∫(tan⁡2x+1)tan⁡46x sec⁡2x dx=∫(u2+1)u46 du=u4949+u4747+C=tan⁡49x49+tan⁡47x47+C\begin{align*} \int\sec^4x\,\tan^{46}x\,\dee{x} &=\int(\tan^2x+1) \tan^{46}x\, \sec^2 x\,\dee{x} =\int (u^2+1)u^{46}\,\dee{u} \\ &=\frac{u^{49}}{49}+\frac{u^{47}}{47}+C =\frac{\tan^{49}x}{49}+\frac{\tan^{47}x}{47}+C \end{align*}
Q14Stage 2

Evaluate ∫tan⁡3xsec⁡1.5x dx\displaystyle\int \tan^3 x \sec^{1.5} x ~\dee{x}.

Hint

Try the substitution u=sec⁡xu=\sec x.

Answer

13.5sec⁡3.5x−11.5sec⁡1.5x+C\dfrac{1}{3.5}\sec^{3.5}x - \dfrac{1}{1.5}\sec^{1.5}x+C

Full solution

We use the substitution u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u} = \sec x \tan x~\dee{x}. Then tan⁡2x=sec⁡2x−1=u2−1\tan^2 x = \sec^2 x - 1 = u^2-1.

∫tan⁡3xsec⁡1.5x dx=∫tan⁡2x⋅sec⁡0.5x⋅sec⁡xtan⁡xdx=∫(u2−1)u0.5 du=∫(u2.5−u0.5) du=u3.53.5−u1.51.5+C=13.5sec⁡3.5x−11.5sec⁡1.5x+C\begin{align*} \int \tan^3 x \sec^{1.5} x ~\dee{x} &= \int \tan^2 x \cdot \sec^{0.5}x \cdot \sec x \tan x \dee{x}\\ &=\int(u^2-1)u^{0.5}~\dee{u} = \int \left(u^{2.5} - u^{0.5}\right)~\dee{u}\\ &=\frac{u^{3.5}}{3.5} - \frac{u^{1.5}}{1.5}+C\\ &=\frac{1}{3.5}\sec^{3.5}x - \frac{1}{1.5}\sec^{1.5}x+C \end{align*}

Note this solution used the same method as Example 1.8.13 in the CLP-2 text for the case that the power of tangent is odd and there is at least one secant.

Q15Stage 2

Evaluate ∫tan⁡3xsec⁡2x dx\displaystyle\int \tan^3x\sec^2x~\dee{x}.

Hint

Compare to Question 14.

Answer

14sec⁡4x−12sec⁡2x+C\dfrac{1}{4}\sec^4 x - \dfrac{1}{2}\sec^2 x +C or 14tan⁡4x+C\dfrac{1}{4}\tan^4 x +C

Full solution

We'll give two solutions.

  • As in Question 14, we have an odd power of tangent and at least one secant. So, as in strategy (2) of Section 1.8.2 in the CLP-2 text, we can use the substitution u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u}=\sec x \tan x~\dee{x}, and tan⁡2x=sec⁡2x−1=u2−1\tan^2 x = \sec^2 x -1=u^2-1.

    ∫tan⁡3xsec⁡2x dx=∫tan⁡2xsec⁡x⋅sec⁡xtan⁡x dx=∫(u2−1)u du=∫(u3−u) du=14u4−12u2+C=14sec⁡4x−12sec⁡2x+C\begin{align*} \int \tan^3x\sec^2x~\dee{x}&=\int \tan^2 x \sec x \cdot \sec x \tan x~\dee{x}\\ &=\int(u^2-1)u~\dee{u} = \int \left(u^3-u\right)~\dee{u}\\ &=\frac{1}{4}u^4 - \frac{1}{2}u^2+C\\ &=\frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 x +C \end{align*}

  • We have an even, strictly positve, power of sec⁡x\sec x. So, as in strategy (3) of Section 1.8.2 in the CLP-2 text, we can use the substitution u=tan⁡xu=\tan x, du=sec⁡2x dx\dee{u}=\sec^2 x~\dee{x}.

    ∫tan⁡3xsec⁡2x dx=∫tan⁡3x⋅sec⁡2x dx=∫u3 du=14u4+C=14tan⁡4x+C\begin{align*} \int \tan^3x\sec^2x~\dee{x}&=\int \tan^3 x \cdot \sec^2 x ~\dee{x}\\ &=\int u^3~\dee{u} \\ &=\frac{1}{4}u^4 + C\\ &=\frac{1}{4}\tan^4x +C \end{align*}

It looks like we have two different answers. But, because tan⁡2x=sec⁡2x−1\tan^2x = \sec^2 x-1,

14tan⁡4=14(sec⁡2x−1)2=14sec⁡4x−12sec⁡2x+14\begin{equation*} \frac{1}{4}\tan^4 = \frac{1}{4} {(\sec^2x -1)}^2 = \frac{1}{4} \sec^4 x - \frac{1}{2}\sec^2 x + \frac{1}{4} \end{equation*}

and the two answers are really the same, except that the arbitrary constant CC of Solution 1 is 14\frac{1}{4} plus the arbitrary constant CC of Solution 2.

Q16Stage 2

Evaluate ∫tan⁡4xsec⁡2x dx\displaystyle\int \tan^4 x \sec^2 x ~\dee{x}.

Hint

What is the derivative of tangent?

Answer

15tan⁡5x+C\dfrac{1}{5}\tan^5 x +C

Full solution

In contrast to Questions 14 and 15, we do not have an odd power of tangent, so we should consider a different substitution. Luckily, if we choose u=tan⁡xu=\tan x, then du=sec⁡2x dx\dee{u}=\sec^2 x~\dee{x}, and this fits our integrand nicely.

∫tan⁡4xsec⁡2x dx=∫u4 du=15u5+C=15tan⁡5x+C\begin{align*} \int \tan^4 x \sec^2 x ~\dee{x}&=\int u^4~\dee{u}=\frac{1}{5}u^5+C = \frac{1}{5}\tan^5 x +C \end{align*}
Q17Stage 2

Evaluate ∫tan⁡3xsec⁡−0.7x dx\displaystyle\int \tan^3 x \sec^{-0.7}x ~\dee{x}.

Hint

Don't be scared off by the non-integer power of secant. You can still use the strategies in the notes for an odd power of tangent.

Answer

11.3sec⁡1.3x+10.7cos⁡0.7x+C\dfrac{1}{1.3}\sec^{1.3}x + \dfrac{1}{0.7}\cos^{0.7}x+C

Full solution

  • Since the power of tangent is odd, let's try to use the substitution u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u} = \sec x \tan x ~\dee{x}, and tan⁡2x=sec⁡2x−1=u2−1\tan^2 x = \sec^2 x -1 = u^2-1, as in Questions 14 and 15. In order to make this work, we need to see sec⁡xtan⁡x dx\sec x \tan x~\dee{x} in the integrand, so we do a little algebraic manipulation.

    ∫tan⁡3xsec⁡−0.7x dx=∫tan⁡3xsec⁡0.7x dx=∫tan⁡3xsec⁡1.7xsec⁡x dx=∫tan⁡2xsec⁡1.7x⋅sec⁡xtan⁡x dx=∫u2−1u1.7 du=∫(u0.3−u−1.7) du=u1.31.3+10.7u0.7+C=11.3sec⁡1.3x+10.7sec⁡0.7x+C=11.3sec⁡1.3x+10.7cos⁡0.7x+C\begin{align*} \int \tan^3 x \sec^{-0.7}x ~\dee{x}&= \int \dfrac{\tan^3 x}{\sec^{0.7 x}}~\dee{x} = \int \dfrac{\tan^3 x}{\sec^{1.7 x}}\sec x~\dee{x}\\ &=\int \frac{\tan^2x}{\sec^{1.7}x}\cdot \sec x \tan x~\dee{x}\\ &=\int \frac{u^2-1}{u^{1.7}}~\dee{u} = \int \left(u^{0.3}-u^{-1.7}\right)~\dee{u}\\ &=\frac{u^{1.3}}{1.3} + \frac{1}{0.7u^{0.7}}+C\\ &=\frac{1}{1.3}\sec^{1.3}x + \frac{1}{0.7\sec^{0.7}x}+C\\ &=\frac{1}{1.3}\sec^{1.3}x + \frac{1}{0.7}\cos^{0.7}x+C \end{align*}

  • Let's convert the secants and tangents to sines and cosines.

    ∫tan⁡3xsec⁡−0.7x dx=∫sin⁡3xcos⁡3x⋅cos⁡0.7x dx=∫sin⁡3xcos⁡2.3x dx=∫sin⁡2xcos⁡2.3x⋅sin⁡x dx\begin{align*}\int \tan^3 x \sec^{-0.7}x ~\dee{x}&= \int \frac{\sin^3 x}{\cos^3 x}\cdot \cos^{0.7}x~\dee{x}\\ &=\int\frac{\sin^3 x}{\cos^{2.3}x}~\dee{x}=\int \frac{\sin^2 x}{\cos^{2.3}x}\cdot\sin x~\dee{x}\end{align*}

    Using the substitution u=cos⁡xu=\cos x, du=−sin⁡ dx\dee{u}=-\sin~\dee{x}, and sin⁡2x=1−cos⁡2x=1−u2\sin^2 x = 1-\cos^2 x = 1-u^2:

    =−∫1−u2u2.3 du=∫(−u−2.3+u−0.3) du=11.3u−1.3+10.7u0.7+C=11.3sec⁡1.3x+10.7cos⁡0.7x+C\begin{align*}& = -\int\frac{1-u^2}{u^{2.3}}~\dee{u} = \int \left(-u^{-2.3}+u^{-0.3}\right)~\dee{u}\\ &=\frac{1}{1.3}u^{-1.3} + \frac{1}{0.7}u^{0.7}+C\\ &=\dfrac{1}{1.3}\sec^{1.3}x + \dfrac{1}{0.7}\cos^{0.7}x+C\end{align*}
Q18Stage 2

Evaluate ∫tan⁡5x dx\displaystyle\int \tan^5 x ~\dee{x}.

Hint

Since there are no secants in the problem, it's difficult to use the substitution u=sec⁡xu=\sec x that we've enjoyed in the past. Example 1.8.12 in the CLP-2 text provides a template for antidifferentiating an odd power of tangent.

Answer

=14sec⁡4x−sec⁡2x+log⁡∣sec⁡x∣+C=\dfrac{1}{4}\sec^4 x - \sec^2 x + \log|\sec x|+C

Full solution

We replace tan⁡x\tan x with sin⁡xcos⁡x\dfrac{\sin x}{\cos x}.

∫tan⁡5x dx=∫(sin⁡xcos⁡x)5 dx=∫sin⁡4xcos⁡5x⋅sin⁡x dx\begin{align*}\int \tan^5 x~\dee{x}&=\int\left(\frac{\sin x}{\cos x}\right)^5~\dee{x} = \int\frac{\sin^4 x}{\cos^5 x}\cdot \sin x~\dee{x}\end{align*}

Now we use the substitution u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x~\dee{x}, and sin⁡2x=1−cos⁡2x=1−u2\sin^2 x = 1-\cos^2 x = 1-u^2.

=−∫(1−u2)2u5 du=∫(−u−5+2u−3−u−1) du=14u−4−u−2−log⁡∣u∣+C=14sec⁡4x−sec⁡2x−log⁡∣cos⁡x∣+C=14sec⁡4x−sec⁡2x+log⁡∣sec⁡x∣+C\begin{align*}&=-\int\frac{(1-u^2)^2}{u^5}~\dee{u} = \int \left(-u^{-5}+2u^{-3}-u^{-1}\right)~\dee{u}\\ &=\frac{1}{4}u^{-4} - u^{-2}-\log|u|+C\\ &=\dfrac{1}{4}\sec^4 x - \sec^2 x - \log|\cos x|+C\\ &=\dfrac{1}{4}\sec^4 x - \sec^2 x + \log|\sec x|+C\end{align*}

where in the last line, we used the logarithm rule log⁡(ba)=alog⁡b\log(b^a) = a\log b, with ba=cos⁡x=(sec⁡x)−1b^a = \cos x = \left(\sec x\right)^{-1}.

Q19Stage 2

Evaluate ∫0π/6tan⁡6x dx\displaystyle\int_0^{\pi/6} \tan^6 x ~\dee{x}.

Hint

Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. You'll want to use the identity tan⁡2x=sec⁡2x−1\tan^2x = \sec^2 x -1, then use the substitution u=tan⁡xu=\tan x, du=sec⁡2x dx\dee{u}=\sec^2 x~\dee{x} on (perhaps only a part of) the resulting integral. Example 1.8.16 in the CLP-2 text show you how this can be accomplished.

Answer

41453−π6\dfrac{41}{45\sqrt{3}} - \dfrac{\pi}{6}

Full solution

Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. For even powers, we use the identity tan⁡2x=sec⁡2x−1\tan^2x = \sec^2 x -1, then use the substitution u=tan⁡xu=\tan x, du=sec⁡2x dx\dee{u}=\sec^2 x~\dee{x} on (perhaps only a part of) the resulting integral.

∫0π/6tan⁡6x dx=∫0π/6tan⁡4x(sec⁡2x−1) dx=∫0π/6(tan⁡4xsec⁡2x⏟u4 du−tan⁡4x) dx=∫0π/6(tan⁡4xsec⁡2x−tan⁡2x(sec⁡2x−1)) dx=∫0π/6(tan⁡4xsec⁡2x−tan⁡2xsec⁡2x⏟u2 du+tan⁡2x) dx=∫0π/6(tan⁡4xsec⁡2x−tan⁡2xsec⁡2x+(sec⁡2x⏟du−1)) dx=∫0π/6(tan⁡4x−tan⁡2x+1)sec⁡2x dx−∫0π/61dx\begin{align*}\int_0^{\pi/6} \tan^6 x ~\dee{x}&=\int_0^{\pi/6} \tan^4 x(\sec^2 x -1) ~\dee{x}\\ &=\int_0^{\pi/6} \bigg(\underbrace{\tan^4 x \sec^2 x}_{u^4~\dee{u}} - \tan^4 x\bigg)~\dee{x}\\ &=\int_0^{\pi/6} \bigg(\tan^4 x \sec^2 x - \tan^2 x(\sec^2 x-1)\bigg)~\dee{x} \\&=\int_0^{\pi/6}\bigg( \tan^4 x \sec^2 x - \underbrace{\tan^2 x\sec^2 x}_{u^2~\dee{u}}+\tan^2 x\bigg)~\dee{x} \\&=\int_0^{\pi/6}\bigg( \tan^4 x \sec^2 x - \tan^2 x\sec^2 x+(\underbrace{\sec^2x}_{\dee{u}}-1)\bigg)~\dee{x} \\&=\int_0^{\pi/6}\left( \tan^4 x - \tan^2 x+1\right)\sec^2 x~\dee{x} - \int_0^{\pi/6} 1\dee{x}\end{align*}

Note tan⁡(0)=0\tan(0)=0, and tan⁡(π/6)=1/3\tan(\pi/6)=1/\sqrt{3}.

=∫01/3(u4−u2+1) du−[x]0π/6=[15u5−13u3+u]01/3−π6=1535−1333+13−π6=41453−π6\begin{align*}&=\int_0^{1/\sqrt{3}}(u^4-u^2+1)~\dee{u} - \big[ x\big]_0^{\pi/6}\\ &=\left[\frac{1}{5}u^5 - \frac{1}{3}u^3+u\right]_0^{1/\sqrt{3}} - \frac{\pi}{6}\\ &=\frac{1}{5\sqrt{3}^5} - \frac{1}{3\sqrt{3}^3}+\frac{1}{\sqrt{3}}-\frac{\pi}{6}\\ &=\dfrac{41}{45\sqrt{3}} - \dfrac{\pi}{6}\end{align*}
Q20Stage 2

Evaluate ∫0π/4tan⁡8xsec⁡4x dx\displaystyle\int_0^{\pi/4} \tan^8 x \sec^4 x ~\dee{x}.

Hint

Since there is an even power of secant in the integrand, we can use the substitution u=tan⁡xu=\tan x.

Answer

111+19\dfrac{1}{11}+\dfrac{1}{9}

Full solution

Since there is an even power of secant in the integrand, we can reserve two secants for du\dee{u} and change the rest to tangents. That sets us up nicely to use the substitution u=tan⁡xu=\tan x, du=sec⁡2x dx\dee{u}=\sec^2 x~\dee{x}. Note tan⁡(0)=0\tan(0)=0 and tan⁡(π/4)=1\tan(\pi/4)=1.

∫0π/4tan⁡8xsec⁡4x dx=∫0π/4tan⁡8x (tan⁡2x+1)sec⁡2x dx=∫01u8 (u2+1) du=∫01u10+u8 du=111+19\begin{align*} \int_0^{\pi/4} \tan^8 x \sec^4 x ~\dee{x}&=\int_0^{\pi/4} \tan^8 x~( \tan^2 x+1) \sec^2 x~\dee{x}\\&=\int_0^{1} u^8 ~( u^2 +1) ~\dee{u}\\ &=\int_0^1 u^{10}+u^8~\dee{u}\\ &=\frac{1}{11}+\frac{1}{9} \end{align*}
Q21Stage 2

Evaluate ∫tan⁡xsec⁡x dx\displaystyle\int \tan x \sqrt{\sec x} ~\dee{x}.

Hint

How have we handled integration in the past that involved an odd power of tangent?

Answer

2sec⁡x+C2\sqrt{\sec x}+C

Full solution

  • Let's use the substitution u=sec⁡xu=\sec x, du=sec⁡xtan⁡x dx\dee{u}=\sec x \tan x~\dee{x}. In order to make this work, we need to see sec⁡xtan⁡x\sec x \tan x in the integrand, so we start with some algebraic manipulation.

    ∫tan⁡xsec⁡x(sec⁡xsec⁡x) dx=∫1sec⁡xsec⁡xtan⁡x dx=∫1u du=2u+C=2sec⁡x+C\begin{align*} \int \tan x \sqrt{\sec x}\left(\frac{\sqrt{\sec x}}{\sqrt{\sec x}}\right) ~\dee{x}&=\int \frac{1}{\sqrt{\sec x}}\sec x\tan x~\dee{x}\\ &=\int \frac{1}{\sqrt{u}}~\dee{u}=2\sqrt{u}+C\\ &=2\sqrt{\sec x}+C \end{align*}

  • Let's turn our secants and tangents into sines and cosines.

    ∫tan⁡xsec⁡x dx=∫sin⁡xcos⁡x⋅cos⁡x dx=∫sin⁡xcos⁡1.5x dx\begin{align*}\int \tan x \sqrt{\sec x}~\dee{x}&=\int \frac{\sin x}{\cos x\cdot\sqrt{\cos x}}~\dee{x}=\int \frac{\sin x}{\cos^{1.5}x}~\dee{x}\end{align*}

    We use the substitution u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x~\dee{x}.

    =∫−u−1.5 du=2u+C=2sec⁡x+C\begin{align*}&=\int -u^{-1.5}~\dee{u}=\frac{2}{\sqrt{u}}+C\\ &=2\sqrt{\sec x}+C\end{align*}
Q22Stage 2

Evaluate ∫sec⁡8θtan⁡eθ dθ\displaystyle\int \sec^{8}\theta \tan^{e}\theta ~\dee{\theta}.

Hint

Remember ee is some constant. What are our strategies when the power of secant is even and positive? We've seen one such substitution in Example 1.8.15 of the CLP-2 text.

Answer

tan⁡e+1θ(tan⁡6θ7+e+3tan⁡4θ5+e+3tan⁡2θ3+e+11+e)+C\tan^{e+1}\theta\left( \dfrac{\tan^{6}\theta}{7+e}+\dfrac{3\tan^4\theta}{5+e}+\dfrac{3\tan^2\theta}{3+e}+\dfrac{1}{1+e} \right)+C

Full solution

Since the power of secant is even and positive, we can reserve two secants for du\dee{u}, and change the rest into tangents, setting the stage for the substitution u=tan⁡θu = \tan \theta, du=sec⁡2θ dθ\dee{u}=\sec^2 \theta~\dee{\theta}.

∫sec⁡8θtan⁡eθ dθ=∫sec⁡6θtan⁡eθsec⁡2θ dθ=∫(tan⁡2θ+1)3tan⁡eθsec⁡2θ dθ=∫(u2+1)3⋅ue du=∫(u6+3u4+3u2+1)⋅ue du=∫(u6+e+3u4+e+3u2+e+ue) du=17+eu7+e+35+eu5+e+33+eu3+e+11+eu1+e+C=17+etan⁡7+eθ+35+etan⁡5+eθ+33+etan⁡3+eθ+11+etan⁡1+eθ+C=tan⁡1+eθ(tan⁡6θ7+e+3tan⁡4θ5+e+3tan⁡2θ3+e+11+e)+C\begin{align*} \int \sec^{8}\theta \tan^{e}\theta ~\dee{\theta}&=\int \sec^6 \theta \tan^e \theta \sec^2 \theta~\dee{\theta}\\ &=\int (\tan^2 \theta +1)^3 \tan^e \theta \sec^2 \theta~\dee{\theta}\\ &=\int (u^2+1)^3 \cdot u^e ~\dee{u}\\ &=\int (u^6+3u^4 +3u^2+1) \cdot u^e ~\dee{u}\\ &=\int (u^{6+e}+3u^{4+e} +3u^{2+e}+ u^e) ~\dee{u}\\ &=\frac{1}{7+e}u^{7+e}+\frac{3}{5+e}u^{5+e}+\frac{3}{3+e}u^{3+e}+\frac{1}{1+e}u^{1+e}+C \\ &=\frac{1}{7+e}\tan^{7+e}\theta+\frac{3}{5+e}\tan^{5+e}\theta+\frac{3}{3+e}\tan^{3+e}\theta+\frac{1}{1+e}\tan^{1+e}\theta+C\\ &=\tan^{1+e}\theta\left( \frac{\tan^{6}\theta}{7+e}+\frac{3\tan^4\theta}{5+e}+\frac{3\tan^2\theta}{3+e}+\frac{1}{1+e} \right)+C \end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q23Stage 3Past exam · 2001D

A reduction formula.

  1. Let nn be a positive integer with n≥2n\ge 2. Derive the reduction formula

    ∫tan⁡n(x) dx=tan⁡n−1(x)n−1−∫tan⁡n−2(x) dx.\int\tan^n(x)\,\dee{x}=\frac{\tan^{n-1}(x)}{n-1} -\int\tan^{n-2}(x)\,\dee{x}.

  2. Calculate ∫0π/4tan⁡6(x) dx\displaystyle\int_0^{\pi/4}\tan^6(x)\,\dee{x}.

Hint

See Example 1.8.16 in the

CLP-2 text for a strategy for integrating powers of tangent.

Answer

(a) Using the trig identity tan⁡2x=sec⁡2x−1\tan^2x=\sec^2 x-1 and the substitution y=tan⁡xy=\tan x, dy=sec⁡2x dx\dee{y}=\sec^2 x\ \dee{x},

∫tan⁡nx dx=∫tan⁡n−2x tan⁡2x dx=∫tan⁡n−2x sec⁡2x dx−∫tan⁡n−2x dx=∫yn−2 dy−∫tan⁡n−2x dx=yn−1n−1−∫tan⁡n−2x dx=tan⁡n−1xn−1−∫tan⁡n−2x dx\begin{alignat*}{3} \int\tan^nx\ \dee{x} &=\int\tan^{n-2}x\ \tan^2x\ \dee{x} &&=\int\tan^{n-2}x\ \sec^2x\ \dee{x}-\int\tan^{n-2}x\ \dee{x}\\ &=\int y^{n-2}\,\dee{y}-\int\tan^{n-2}x\ \dee{x} &&=\frac{y^{n-1}}{n-1}-\int\tan^{n-2}x\ \dee{x}\\ &=\frac{\tan^{n-1}x}{n-1} -\int\tan^{n-2}x\ \dee{x} \end{alignat*}

(b) 1315−π4≈0.0813\displaystyle\frac{13}{15}-\frac{\pi}{4}\approx0.0813

Full solution

(a) Using the trig identity tan⁡2x=sec⁡2x−1\tan^2x=\sec^2 x-1 and the substitution y=tan⁡xy=\tan x, dy=sec⁡2x dx\dee{y}=\sec^2 x\ \dee{x},

∫tan⁡nx dx=∫tan⁡n−2x tan⁡2x dx=∫tan⁡n−2x sec⁡2x dx−∫tan⁡n−2x dx=∫yn−2 dy−∫tan⁡n−2x dx=yn−1n−1−∫tan⁡n−2x dx=tan⁡n−1xn−1−∫tan⁡n−2x dx\begin{alignat*}{3} \int\tan^nx\ \dee{x} &=\int\tan^{n-2}x\ \tan^2x\ \dee{x} &&=\int\tan^{n-2}x\ \sec^2x\ \dee{x}-\int\tan^{n-2}x\ \dee{x}\\ &=\int y^{n-2}\,\dee{y}-\int\tan^{n-2}x\ \dee{x} &&=\frac{y^{n-1}}{n-1}-\int\tan^{n-2}x\ \dee{x}\\ &=\frac{\tan^{n-1}x}{n-1} -\int\tan^{n-2}x\ \dee{x} \end{alignat*}

(b) By the reduction formula of part (a),

∫0π/4tan⁡n(x) dx=[tan⁡n−1xn−1]0π/4−∫0π/4tan⁡n−2(x) dx=1n−1−∫0π/4tan⁡n−2(x) dx\begin{align*} \int_0^{\pi/4}\tan^n(x)\,\dee{x}&= \left[\frac{\tan^{n-1}x}{n-1}\right]_{0}^{\pi/4}-\int_0^{\pi/4}\tan^{n-2}(x)\,\dee{x}\\ &=\frac{1}{n-1}-\int_0^{\pi/4}\tan^{n-2}(x)\,\dee{x} \end{align*}

for all integers n≥2n\ge 2, since tan⁡0=0\tan 0=0 and tan⁡π4=1\tan\frac{\pi}{4}=1. We apply this reduction formula, with n=6,4,2n=6,4,2.

∫0π/4tan⁡6(x) dx=15−∫0π/4tan⁡4(x) dx=15−13+∫0π/4tan⁡2(x) dx=15−13+1−∫0π/4 dx=15−13+1−π4=1315−π4\begin{align*} \int_0^{\pi/4}\tan^6(x)\,\dee{x} &=\frac{1}{5}-\int_0^{\pi/4}\tan^4(x)\,\dee{x} =\frac{1}{5}-\frac{1}{3}+\int_0^{\pi/4}\tan^2(x)\,\dee{x} =\frac{1}{5}-\frac{1}{3}+1-\int_0^{\pi/4}\,\dee{x}\cr &=\frac{1}{5}-\frac{1}{3}+1-\frac{\pi}{4} =\frac{13}{15}-\frac{\pi}{4} \end{align*}

Using a calculator, we see this is approximately 0.08130.0813.

Notice how much faster this was than the method of Question 19.

Q24Stage 3

Evaluate ∫tan⁡5xcos⁡2x dx\displaystyle\int \tan^5 x \cos^2 x ~\dee{x}.

Hint

Write tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}.

Answer

12cos⁡2x+2log⁡∣cos⁡x∣−12cos⁡2x+C\dfrac{1}{2\cos^2 x}+2\log|\cos x|-\dfrac{1}{2}\cos^2 x +C

Full solution

Recall tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}.

∫tan⁡5xcos⁡2x dx=∫sin⁡5xcos⁡5xcos⁡2x dx=∫sin⁡5xcos⁡3x dx\begin{align*}\int \tan^5 x \cos^2 x ~\dee{x}&=\int \frac{\sin^5 x}{\cos^5 x}\cos^2 x ~\dee{x} =\int \frac{\sin^5 x}{\cos^3 x}~\dee{x}\end{align*}

Substitute u=cos⁡xu=\cos x, so du=−sin⁡x dx\dee{u}=-\sin x~\dee{x} and sin⁡2x=1−cos⁡2x=1−u2\sin^2 x = 1-\cos^2 x = 1-u^2.

=∫sin⁡4xcos⁡3x sin⁡x dx=−∫(1−u2)2u3 du=−∫1−2u2+u4u3 du=∫(−1u3+2u−u)du=12u2+2log⁡∣u∣−12u2+C=12cos⁡2x+2log⁡∣cos⁡x∣−12cos⁡2x+C\begin{align*}&=\int \frac{\sin^4 x}{\cos^3 x}~\sin x~\dee{x} =-\int \frac{(1-u^2)^2}{u^3}~\dee{u}\\ &=-\int \frac{1-2u^2+u^4}{u^3}~\dee{u}= \int \left(-\frac{1}{u^3}+\frac{2}{u}-u\right)\dee{u}\\ &=\frac{1}{2u^2}+2\log|u|-\frac{1}{2}u^2+C\\ &=\frac{1}{2\cos^2 x}+2\log|\cos x|-\frac{1}{2}\cos^2 x +C\end{align*}
Q25Stage 3

Evaluate ∫1cos⁡2θdθ\displaystyle\int \frac{1}{\cos^2 \theta}\dee{\theta}.

Hint

1cos⁡θ=sec⁡θ\dfrac{1}{\cos \theta} = \sec \theta

Answer

tan⁡θ+C\tan \theta +C

Full solution

We can use the definition of secant to make this integral look more familiar.

∫1cos⁡2θ dθ=∫sec⁡2θ dθ=tan⁡θ+C\int \frac{1}{\cos^2 \theta}~\dee{\theta} = \int \sec^2\theta~\dee{\theta} = \tan \theta +C
Q26Stage 3

Evaluate ∫cot⁡x dx\displaystyle\int \cot x~\dee{x}.

Hint

cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{\cos x}{\sin x}

Answer

log⁡∣sin⁡x∣+C\log|\sin x|+C

Full solution

We re-write cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{\cos x}{\sin x}, and use the substitution u=sin⁡xu=\sin x, du=cos⁡x dx\dee{u}=\cos x~\dee{x}.

∫cot⁡x dx=∫cos⁡xsin⁡x dx=∫1u du=log⁡∣u∣+C=log⁡∣sin⁡x∣+C\begin{align*} \int \cot x~\dee{x}&= \int \frac{\cos x}{\sin x}~\dee{x} = \int \frac{1}{u}~\dee{u}\\ &=\log|u|+C = \log|\sin x|+C \end{align*}
Q27Stage 3

Evaluate ∫exsin⁡(ex)cos⁡(ex) dx\displaystyle\int e^x\sin(e^x)\cos(e^x) ~\dee{x}.

Hint

Try substituting.

Answer

12sin⁡2(ex)+C\dfrac{1}{2}\sin^2(e^x)+C

Full solution

  • We begin with the obvious substitution, w=exw=e^x, dw=exdw\dee{w}=e^x \dee{w}.

    ∫exsin⁡(ex)cos⁡(ex) dx=∫sin⁡wcos⁡w dw\begin{align*}\int e^x\sin(e^x)\cos(e^x) ~\dee{x}&= \int \sin w \cos w ~\dee{w}\end{align*}

    Now we see another substitution, u=sin⁡wu=\sin w, du=cos⁡w dw\dee{u}=\cos w~\dee{w}.

    =∫u du=12u2+C=12sin⁡2w+C=12sin⁡2(ex)+C\begin{align*}&=\int u~\dee{u}=\frac{1}{2}u^2+C=\frac{1}{2}\sin^2 w +C\\ &=\frac{1}{2}\sin^2(e^x)+C\end{align*}

  • Notice that ddx{sin⁡(ex)}=excos⁡(ex)\diff{}{x}\{\sin(e^x)\} = e^x \cos(e^x). This suggests to us the substitution u=sin⁡(ex)u=\sin(e^x), du=excos⁡(ex) dx\dee{u} = e^x \cos(e^x)~\dee{x}.

    ∫exsin⁡(ex)cos⁡(ex) dx=∫u du=12u2+C=12sin⁡2(ex)+C\begin{align*} \int e^x\sin(e^x)\cos(e^x) ~\dee{x}&= \int u~\dee{u} =\frac{1}{2}u^2+C = \frac{1}{2}\sin^2(e^x)+C \end{align*}
Q28Stage 3

Evaluate ∫sin⁡(cos⁡x)sin⁡3x dx\displaystyle\int \sin(\cos x)\sin^3 x ~\dee{x}.

Hint

To deal with the “inside function," start with a substitution.

Answer

(sin⁡2x+2)cos⁡(cos⁡x)+2cos⁡xsin⁡(cos⁡x)+C(\sin^2x+2)\cos (\cos x) + 2\cos x\sin (\cos x) +C

Full solution

Since we have an “inside function," we start with the substitution s=cos⁡xs=\cos x, so −ds=sin⁡x dx-\dee{s}=\sin x ~\dee{x} and sin⁡2x=1−cos⁡2x=1−s2\sin^2 x = 1-\cos^2 x = 1-s^2.

∫sin⁡(cos⁡x)sin⁡3x dx=∫sin⁡(cos⁡x)⋅sin⁡2x⋅sin⁡xdx=−∫sin⁡(s)⋅(1−s2) ds\begin{align*}\int \sin(\cos x)\sin^3 x ~\dee{x}&=\int \sin(\cos x) \cdot \sin^2 x \cdot \sin x \dee{x}\\ &=-\int \sin(s)\cdot (1-s^2) ~\dee{s}\end{align*}

We use integration by parts with u=(1−s2)u=(1-s^2), dv=sin⁡s ds\dee{v}=\sin s ~\dee{s}; du=−2s ds\dee{u}=-2s~\dee{s}, and v=−cos⁡sv = -\cos s.

=−[−(1−s2)cos⁡s−∫2scos⁡s ds]=(1−s2)cos⁡s+∫2scos⁡s ds\begin{align*}&=-\left[-(1-s^2)\cos s - \int 2s\cos s~ \dee{s}\right] \\&= (1-s^2)\cos s + \int 2s\cos s~ \dee{s}\end{align*}

We integrate by parts again, with u=2su=2s, dv=cos⁡s ds\dee{v}=\cos s ~\dee{s}; du=2 ds\dee{u}=2~\dee{s}, and v=sin⁡sv=\sin s.

=(1−s2)cos⁡s+2ssin⁡s−∫2sin⁡s ds=(1−s2)cos⁡s+2ssin⁡s+2cos⁡s+C=sin⁡2x⋅cos⁡(cos⁡x)+2cos⁡x⋅sin⁡(cos⁡x)+2cos⁡(cos⁡x)+C=(sin⁡2x+2)cos⁡(cos⁡x)+2cos⁡x⋅sin⁡(cos⁡x)+C\begin{align*}&=(1-s^2)\cos s + 2s\sin s - \int 2\sin s~\dee{s} \\&=(1-s^2)\cos s + 2s\sin s +2\cos s +C \\&=\sin^2 x\cdot\cos (\cos x) + 2\cos x\cdot\sin (\cos x) +2\cos (\cos x) +C \\&=(\sin^2x+2)\cos (\cos x) + 2\cos x\cdot\sin (\cos x) +C\end{align*}
Q29Stage 3

Evaluate ∫xsin⁡xcos⁡x dx\displaystyle\int x\sin x \cos x ~\dee{x}.

Hint

Try an integration by parts.

Answer

x2sin⁡2x−x4+14sin⁡xcos⁡x+C\dfrac{x}{2}\sin^2 x - \dfrac{x}{4} +\dfrac{1}{4}\sin x \cos x+C

Full solution

Since the integrand is the product of polynomial and trigonometric functions, we suspect it might yield to integration by parts. There are a number of ways this can be accomplished.

  • Before we choose parts, let's use the identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x.

    ∫xsin⁡xcos⁡xdx=12∫xsin⁡(2x)dx\begin{align*}\int x\sin x \cos x \dee{x}&=\frac{1}{2}\int x \sin(2x)\dee{x}\end{align*}

    Now let u=xu= x, dv=sin⁡(2x)dx\dee{v}=\sin(2x)\dee{x}; du=dx\dee{u}=\dee{x}, and v=−12cos⁡(2x)v=-\frac{1}{2}\cos (2x). Using integration by parts:

    =12[−x2cos⁡(2x)+12∫cos⁡(2x)dx]=−x4cos⁡(2x)+18sin⁡(2x)+C=−x4(1−2sin⁡2x)+14sin⁡xcos⁡x+C=−x4+x2sin⁡2x+14sin⁡xcos⁡x+C\begin{align*}&=\frac{1}{2}\left[-\frac{x}{2}\cos (2x) +\frac{1}{2} \int \cos (2x) \dee{x}\right]\\ &=-\frac{x}{4}\cos (2x) +\frac{1}{8}\sin (2x) +C\\ &=-\frac{x}{4}(1-2\sin^2x) +\frac{1}{4}\sin x\cos x +C\\ &=-\frac{x}{4} + \frac{x}{2}\sin^2x+\frac{1}{4}\sin x \cos x +C\end{align*}

  • If we let u=xu=x, then du=dx\dee{u}=\dee{x}, and this seems desirable for integration by parts. If u=xu=x, then dv=sin⁡xcos⁡xdx\dee{v} = \sin x \cos x \dee{x}. To find vv we can use the substitution u=sin⁡xu=\sin x, du=cos⁡xdx\dee{u}=\cos x \dee{x}.

    v=∫sin⁡xcos⁡xdx=∫udu=12u2+C=12sin⁡2x+C\begin{align*}v=\int \sin x \cos x \dee{x}&=\int u \dee{u} = \frac{1}{2}u^2+C = \frac{1}{2}\sin^2 x +C\end{align*}

    So, we take v=12sin⁡2xv = \frac{1}{2}\sin^2 x. Now we can apply integration by parts to our original integral.

    ∫xsin⁡xcos⁡x dx=x2sin⁡2x−∫12sin⁡2xdx\begin{align*}\int x\sin x \cos x ~\dee{x}&=\frac{x}{2}\sin^2 x - \int \frac{1}{2}\sin^2 x \dee{x}\end{align*}

    Apply the identity sin⁡2x=1−cos⁡(2x)2\sin^2x = \dfrac{1-\cos(2x)}{2}.

    =x2sin⁡2x−14∫1−cos⁡(2x)dx=x2sin⁡2x−x4+18sin⁡(2x)+C=x2sin⁡2x−x4+14sin⁡xcos⁡x+C\begin{align*}&=\frac{x}{2}\sin^2 x - \frac{1}{4}\int 1-\cos(2 x) \dee{x}\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} +\frac{1}{8}\sin(2 x)+C\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} +\frac{1}{4}\sin x \cos x+C\end{align*}

  • Let u=xsin⁡xu=x\sin x and dv=cos⁡xdx\dee{v}=\cos x \dee{x}; then du=(xcos⁡x+sin⁡x)dx\dee{u} = (x\cos x + \sin x)\dee{x} and v=sin⁡xv = \sin x.

    ∫xsin⁡xcos⁡xdx=xsin⁡2x−∫sin⁡x(xcos⁡x+sin⁡x)dx=xsin⁡2x−∫xsin⁡xcos⁡xdx−∫sin⁡2xdx\begin{align*}\int x\sin x \cos x \dee{x}&=x\sin^2 x - \int \sin x(x\cos x + \sin x)\dee{x}\\ &=x\sin^2 x - \int x \sin x \cos x\dee{x} - \int \sin^2 x\dee{x}\end{align*}

    Apply the identity sin⁡2x=1−cos⁡(2x)2\sin^2 x = \dfrac{1-\cos(2x)}{2} to the second integral.

    =xsin⁡2x−∫xsin⁡xcos⁡xdx−∫1−cos⁡(2x)2dx=xsin⁡2x−∫xsin⁡xcos⁡xdx−x2+14sin⁡(2x)+C\begin{align*}&=x\sin^2 x - \int x \sin x \cos x\dee{x} - \int \dfrac{1-\cos(2x)}{2}\dee{x} \\&=x\sin^2 x - \int x \sin x \cos x\dee{x} - \frac{x}{2} +\frac{1}{4}\sin(2x)+C\end{align*}

    So, we have the equation

    ∫xsin⁡xcos⁡xdx=xsin⁡2x−∫xsin⁡xcos⁡xdx−x2+14sin⁡(2x)+C2∫xsin⁡xcos⁡xdx=xsin⁡2x−x2+14sin⁡(2x)+C∫xsin⁡xcos⁡xdx=x2sin⁡2x−x4+18sin⁡(2x)+C2=x2sin⁡2x−x4+14sin⁡xcos⁡x+C2\begin{align*}\color{red}\int x\sin x \cos x \dee{x}&=x\sin^2 x -\textcolor{red}{ \int x \sin x \cos x\dee{x}} - \frac{x}{2} + \frac{1}{4}\sin(2x)+C\\ \color{red}2\int x\sin x \cos x \dee{x}&=x\sin^2 x - \frac{x}{2} + \frac{1}{4}\sin(2x)+C\\ \int x\sin x \cos x \dee{x}&=\frac{x}{2}\sin^2 x - \frac{x}{4} + \frac{1}{8}\sin(2x)+\frac{C}{2}\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} + \frac{1}{4}\sin x\cos x+\frac{C}{2}\end{align*}

    Since CC is an arbitrary constant that can take any number in (−∞,∞)(-\infty,\infty), also C2\frac{C}{2} is an arbitrary constant that can take any number in (−∞,∞)(-\infty,\infty), so we're free to rename C2\frac{C}{2} to CC.

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.