1.8 Trigonometric Integrals

Go ahead and try it!

(e)

If $u=\cos x$, then $\dee{u}=-\sin x\,\dee{x}$. If $n \neq -1$, then

If $n=-1$, then

So, (e) $n$ can be any real number.

Use the substitution $u=\sec x$.

$\dfrac{1}{n}\sec^n x +C$

We use the substitution $u=\sec x$, $\dee{u}=\sec x \tan x ~\dee{x}$.

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

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

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

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

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).

$\sin x-\dfrac{\sin^3 x}{3} +C$

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

See Example 1.8.7 in the

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

$\dfrac{\pi}{2}$

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

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

$\dfrac{\sin^{37}t}{37}-\dfrac{\sin^{39}t}{39}+C$

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

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

$\dfrac{1}{3\cos^3 x} - \dfrac{1}{\cos x}+C$

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

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.

$\displaystyle\frac{\pi}{8} -\frac{9\sqrt3}{64}$

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

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

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

$-\cos x + \dfrac{2}{3}\cos^3 x - \dfrac{1}{5}\cos^5 x +C$

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

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

$\dfrac{1}{2.2}\sin^{2.2}x+C$

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

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.

Try a substitution.

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

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

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

$\dfrac{1}{7}\sec^7 x -\dfrac{1}{5}\sec^5 x + C$

A substitution will work. See Example 1.8.14 in the

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

$\displaystyle\frac{\tan^{49}x}{49}+\frac{\tan^{47}x}{47}+C$

Use the substitution $u=\tan x$, so that $\dee{u}=\sec^2 x\,\dee{x}$:

Try the substitution $u=\sec x$.

$\dfrac{1}{3.5}\sec^{3.5}x - \dfrac{1}{1.5}\sec^{1.5}x+C$

We use the substitution $u=\sec x$, $\dee{u} = \sec x \tan x~\dee{x}$. Then $\tan^2 x = \sec^2 x - 1 = u^2-1$.

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.

Compare to Question 14.

$\dfrac{1}{4}\sec^4 x - \dfrac{1}{2}\sec^2 x +C$ or $\dfrac{1}{4}\tan^4 x +C$

We'll give two solutions.

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

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

What is the derivative of tangent?

$\dfrac{1}{5}\tan^5 x +C$

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 $u=\tan x$, then $\dee{u}=\sec^2 x~\dee{x}$, and this fits our integrand nicely.

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.

$\dfrac{1}{1.3}\sec^{1.3}x + \dfrac{1}{0.7}\cos^{0.7}x+C$

Since there are no secants in the problem, it's difficult to use the substitution $u=\sec x$ 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.

$=\dfrac{1}{4}\sec^4 x - \sec^2 x + \log|\sec x|+C$

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

Now we use the substitution $u=\cos x$, $\dee{u}=-\sin x~\dee{x}$, and $\sin^2 x = 1-\cos^2 x = 1-u^2$.

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

Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. You'll want to use the identity $\tan^2x = \sec^2 x -1$, then use the substitution $u=\tan x$, $\dee{u}=\sec^2 x~\dee{x}$ 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.

$\dfrac{41}{45\sqrt{3}} - \dfrac{\pi}{6}$

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

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

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

$\dfrac{1}{11}+\dfrac{1}{9}$

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

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

$2\sqrt{\sec x}+C$

Remember $e$ 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.

$\tan^{e+1}\theta\left( \dfrac{\tan^{6}\theta}{7+e}+\dfrac{3\tan^4\theta}{5+e}+\dfrac{3\tan^2\theta}{3+e}+\dfrac{1}{1+e} \right)+C$

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