1.13 More Integration Examples

Each option in each column should be used exactly once.

(A)–(I), (B)–(IV), (C)–(II), (D)–(III)

The integrand is the product of sines and cosines. See how this was handled with a substitution in Section 1.8.1 of the CLP-2 text.

After your substitution, you should have a polynomial expression in $u$–but it might take some simplification to get it into a form you can easily integrate.

$\dfrac{1}{5}-\dfrac{2}{7}+\dfrac{1}{9} =\dfrac{8}{315}$

We notice that the integrand has a quadratic polynomial under the square root. If that polynomial were a perfect square, we could get rid of the square root: try a trig substitution, as in Section 1.9 of the CLP-2 text.

The identity $\sin(2\theta)=2\sin\theta\cos\theta$ might come in handy.

$\dfrac{3}{2\sqrt{5}} \arcsin\left(x\sqrt{\dfrac{5}{3}}\right) + \dfrac{x}{2}\sqrt{3-5x^2} +C$

Notice the integral is improper. When you compute the limit, l'H^opital's rule might help.

If you're struggling to think of how to antidifferentiate, try writing $\dfrac{x-1}{e^x} = (x-1)e^{-x}$.

0

Which method usually works for rational functions (the quotient of two polynomials)?

$\log\left|\dfrac{x+1}{3x+1}\right|+C$

It would be nice to replace logarithm with its derivative, $\dfrac{1}{x}$.

$\dfrac{8}{3}\log2-\dfrac{7}{9}$

The integrand is a rational function, so it is possible to use partial fractions. But there is a much easier way!

$\dfrac{1}{2}\log\big|x^2-3\big| + C$

You should prepare your own personal internal list of integration techniques ordered from easiest to hardest. You should have associated to each technique your own personal list of signals that you use to decide when the technique is likely to be useful.

(a) $2$ (b) $\dfrac{2}{15}$ (c) $\dfrac{3e^4}{16}+\dfrac{1}{16}$

Despite both containing a trig function, the two integrals are easiest to evaluate using different methods.

(a) $1$ (b) $\dfrac{8}{15}$

For the integral of secant, see See Section 1.8.3 or Example 1.10.5 in the

CLP-2 text.

In (c), notice the denominator is not yet entirely factored.

(a) $e^2+1$ (b) $\log(\sqrt{2}+1)$ (c) $\log\frac{15}{13}\approx 0.1431$

Part (a) can be done by inspection – use a little highschool geometry! Part (b) is reminiscent of the antiderivative of logarithm–how did we find that one out? Part (c) is an improper integral.

(a) $\dfrac{9}{4}\pi$ (b) $\log 2-2+\dfrac{\pi}{2}\approx 0.264$ (b) $2\log 2-\half\approx0.886$

Use the substitution $u=\sin\theta$.

$\displaystyle\frac{1}{3}\sin^3\theta-2\sin\theta+12\log\left|\frac{\sin\theta-3}{\sin\theta-2}\right|+C$

For (c), try a little algebra to split the integral into pieces that are easy to antidifferentiate.

(a) $\dfrac{1}{15}$ (b) $\dfrac{1}{9}\cdot\dfrac{x}{\sqrt{x^2+9}}+C$ (c) $\dfrac{1}{2}\log|x-1|-\dfrac{1}{4}\log(x^2+1)-\dfrac{1}{2}\arctan x+C$

(d) $\dfrac12\big[ x^2\arctan x -x +\arctan x\big]+C$

If you're stumped, review Sections 1.8, 1.9, and 1.10 in the CLP-2 text.

(a) $\dfrac{1}{12}$ (b) $\displaystyle2\sin^{-1}\frac{x}{2}+x\sqrt{1-\frac{x^2}{4}}+C$

(c) $\displaystyle-2\log|x|+\frac{1}{x}+2\log|x-1|+C$

For part (a), see Example 1.7.11 in the

CLP-2 text. For part (d), see Example 1.10.4 in the

CLP-2 text.

(a) $\displaystyle\frac{2}{5}$ (b) $\displaystyle\frac{1}{2\sqrt{2}}$ (c) $\displaystyle\log 2-\frac{1}{2}\approx 0.193$ (d) $\displaystyle\log 2-\frac{1}{2}\approx0.193$

For part (b), first complete the square in the denominator. You can save some work by first comparing the derivative of the denominator with the numerator. For part (d) use a simple substitution.

(a) $\displaystyle\frac12 x^2\log x -\frac{1}{4} x^2+C$ (b) $\displaystyle\frac{1}{2} \log [x^2+4x+5] -3\arctan(x+2)+C$

(c) $\displaystyle\frac12\log|x-3|-\frac12\log|x-1|+C$ (d) $\displaystyle\frac{1}{3}\arctan x^3 +C$

For part (b), complete the square in the denominator. You can save some work by first comparing the derivative of the denominator with the numerator.

(a) $\displaystyle\frac{\pi}{4}-\frac{1}{2}\log 2$ (b) $\displaystyle\log |x^2-2x+5| +\frac{1}{2}\arctan \frac{x-1}{2} +C$

For part (a), the numerator is the derivative of a function that appears in the denominator.

(a) $\displaystyle-\frac{1}{300(x^3+1)^{100}} + C$ (b) $\displaystyle\frac{\sin^5\!x}{5}-\frac{\sin^7\!x}{7}+C$

The integral is improper.

-2

For part (a), can you convert this into a partial fractions integral? For part (b), start by completing the square inside the square root.

(a) $\displaystyle-\frac{1}{4}\log|e^x+1| +\frac{1}{4}\log|e^x-3| + C$ (b) $\displaystyle\frac{4\pi}{3}-2\sqrt{3}$

For part (b), the numerator is the derivative of a function that is embedded in the denominator.

(a) $\displaystyle\frac{1}{2}\sec^2 x+\log|\cos x|+C$ (b) $\displaystyle \frac{1}{10}\arctan 8\approx 0.1446$

Try a substitution.

$\displaystyle\frac{2}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2}+C$

Note the quadratic function under the square root: you can solve this with trigonometric substitution, as in Section 1.9 of the CLP-2 text.

$\log \left| x+{\sqrt{x^2-2}} \right|-\dfrac{\sqrt{x^2-2}}{x}+C$

Try a substitution, as in Section 1.8.2 of the CLP-2 text.

$\dfrac{7}{24}$

What's the usual trick for evaluating a rational function (quotient of polynomials)?

$3\log|x+1|+\dfrac{2}{x+1}-\dfrac{5}{2(x+1)^2}+C$

If the denominator were $x^2+1$, the antiderivative would be arctangent.

$\dfrac{2}{\sqrt3}\arctan\left(\dfrac{2}{\sqrt3}x+\dfrac{1}{\sqrt3}\right) +C$

Simplify first.

$\displaystyle\frac12\left(x -\sin x \cos x \right)+C$

$x^3+1 = (x+1)(x^2-x+1)$

$\displaystyle\frac{1}{3}\log|x+1| - \frac{1}{6}\log|x^2+x+1| + \frac{1}{\sqrt3}\arctan\left(\frac{2x-1}{\sqrt3}\right)+C$

You have the product of two quite dissimilar functions in the integrand–try integration by parts.

$3x^3\arcsin x + 3\sqrt{1-x^2}-(1-x^2)^{3/2} + C$

Use the identity $\cos(2x) = 2\cos^2x -1$.

2

Using logarithm rules can make the integrand simpler.

$\dfrac{1}{4}$

What is the derivative of the function in the denominator? How could that be useful to you?

$\log\left(\dfrac{\log(\cos(0.1))}{\log(\cos(0.2))}\right)$

For part (a), the substitution $u=\log x$ gives an integral that you have seen before.

(a) $\displaystyle \frac{1}{2} x \big[\sin(\log x) -\cos(\log x)\big]+C$ (b) $2\log 2 -\log 3 = \log\frac{4}{3}$

For part (a), split the integral in two. One part may be evaluated by interpreting it geometrically, without doing any integration at all. For part (c), multiply both the numerator and denominator by $e^x$ and then make a substitution.

(a) $\displaystyle\frac{9}{4}\pi + 9$ (b) $\displaystyle2 \log |x-2| - \log (x^2 + 4) + C$ (c) $\displaystyle\frac{\pi}{2}$

Let $u=\sqrt{1-x}$.

$-\arcsin (\sqrt{1-x})-\sqrt{1-x}\sqrt{x}+C$

Use the substitution $u=e^x$.

$e^e(e-1)$

Use integration by parts. If you choose your parts well, the resulting integration will be very simple.

$\displaystyle\frac{e^x}{x+1}+C$

$\frac{\sin x}{\cos^2 x} = \tan x \sec x$

$\displaystyle x\sec x - \log|\sec x + \tan x|+C$

The cases $n=-1$ and $n=-2$ are different from all other values of $n$.

$\displaystyle\int x(x+a)^n~\dee{x}=\begin{cases} \frac{(x+a)^{(n+2)}}{n+2}-a\frac{(x+a)^{n+1}}{n+1}+C&\text{ if } n \neq -1,-2\\ (x+a)-a\log|x+a|+C & \text{ if } n=-1\\ \log|x+a| + \frac{a}{x+a}+C& \text{ if } n=-2 \end{cases}$

$x^4+1 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$

$x\arctan(x^2) - \frac{1}{\sqrt 2}\Bigg( \frac{1}{2}\log\left|\tfrac{x^2-\sqrt2x+1}{ x^2+\sqrt2x+1}\right| +\arctan\left(\sqrt{2}x+1\right)+ \arctan\left(\sqrt{2}x-1\right) \Bigg)+C$