1.9 Trigonometric Substitution

The beginning of this section has a template for choosing a substitution. Your goal is to use a trig identity to turn the argument of the square root into a perfect square, so you can cancel $\sqrt{(\text{something})^2}=|\text{something}|$.

(a) $x=\dfrac{4}{3}\sec\theta$ (b) $x=\dfrac{1}{2}\sin\theta$ (c) $x=5\tan\theta$

You want to do the same thing you did in Question 1, but you'll have to complete the square first.

(a) $x-2=\sqrt{3}\sec u$ (b) $x-1=\sqrt{5}\sin u$ (c) $\left(2x+\dfrac{3}{2}\right) =\dfrac{\sqrt{31}}{2}\tan u$

(d) $x - \dfrac{1}{2}=\dfrac{1}{2}\sec u$

Since $\theta$ is acute, you can draw it as an angle of a right triangle. The given information will let you label two sides of the triangle, and the Pythagorean Theorem will lead you to the third.

(a) $\dfrac{\sqrt{399}}{20}$ (b) $\dfrac{5\sqrt{2}}{7}$ (c) $\dfrac{\sqrt{x-5}}{2}$

You can draw a right triangle with angle $\theta$, and use the given information to label two of the sides. The Pythagorean Theorem gives you the third side.

(a) $\dfrac{\sqrt{4-x^2}}{2}$ (b) $\dfrac{1}{2}$ (c) $\dfrac{1}{\sqrt{1-x}}$

As in Question 1, choose an appropriate substitution. Your answer should be in terms of your original variable, $x$, which can be achieved using the methods of Question 3.

$\dfrac14\cdot \dfrac x{\sqrt{x^2+4}} + C$

As in Question 1, choose an appropriate substitution. Your answer will be a number, so as long as you change your limits of integration when you substitute, you don't need to bother changing the antiderivative back into the original variable $x$. However, you might want to use the techniques of Question 4 to simplify your final answer.

$\dfrac{1}{2\sqrt{5}}$

Question 1 guides the way to finding the appropriate substitution. Since the integral is definite, your final answer will be a number. Your limits of integration should be common reference angles.

$\dfrac{\pi}{6}$

Question 1 guides the way to finding the appropriate substitution. Since you have in indefinite integral, make sure to get your answer back in terms of the original variable, $x$. Question 3 gives a reliable method for this.

$\displaystyle\log\left|\sqrt{1+\frac{x^2}{25}}+\frac{x}{5}\right|+C$

A trig substitution is not the easiest path.

$\dfrac{1}{2}\sqrt{2x^2+4x}+C$

To antidifferentiate, change your trig functions into sines and cosines.

$-\displaystyle\frac{1}{16}\dfrac{\sqrt{x^2+16}}{x}+C$

The integrand should simplify quite far after your substitution.

$\displaystyle\frac{\sqrt{x^2-9}}{9x} +C$

In part (a) you are asked to integrate an even power of $\cos x$. For part (b) you can use a trigonometric substitution to reduce the integral of part (b) almost to the integral of part (a).

(a) We'll use the trig identity $\cos2\theta=2\cos^2\theta-1$. It implies that

So,

as required.

(b) $\dfrac{8+3\pi}{16}$

What is the symmetry of the integrand?

0

See Example 1.9.3 in the CLP-2 text.

$\displaystyle2\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{4-x^2}+ C$

To integrate an even power of tangent, use the identity $\tan^2 x = \sec^2 x - 1$.

$\sqrt{25x^2-4}-2\arcsec\frac{5x}{2} + C$

A trig substitution is not the easiest path.

$\dfrac{40}{3}$

Complete the square. Your final answer will have an inverse trig function in it.

$\arcsin\dfrac{x+1}{2} + C$

To antidifferentiate even powers of cosine, use the formula $\cos^2\theta = \frac{1}{2}(1+\cos(2\theta))$. Then, remember $\sin(2\theta)=2\sin\theta\cos\theta$.

$\displaystyle\frac{1}{4}\left(\arccos\left(\frac{1}{2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C$, or equivalently,
$\displaystyle\frac{1}{4}\left(\arcsec\left({2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C$

After substituting, use the identity $\tan^2 x = \sec^2 x - 1$ more than once.
Remember $\displaystyle\int \sec x \dee{x} = \log \big|\sec x + \tan x \big|+C$.

$\log(1+\sqrt{2})-\dfrac{1}{\sqrt{2}}$

There's no square root, but we can still make use of the substitution $x=\tan\theta$.

$\displaystyle\frac{1}{2}\left(\arctan x + \frac{x}{x^2+1}\right)+C$

You'll probably want to use the identity $\tan^2\theta+1=\sec^2\theta$ more than once.

$\dfrac{3+x}{2}\sqrt{x^2-2x+2}+ \dfrac{1}{2}\log\left|\sqrt{x^2-2x+2}+x-1\right|+C$

Complete the square — refer to Question 2 if you want a refresher. The constants aren't pretty, but don't let them scare you.

$\displaystyle\frac{1}{\sqrt{3}}\log\left| \left(\frac{6}{5}x+1\right)+\frac{2}{5}\sqrt{9x^2+15x} \right|+C$

After substituting, use the identity $\sec^2 u = \tan^2 u +1$. It might help to break the integral into a few pieces.

$\dfrac{1}{3}\sqrt{1+x^2}(4+x^2)+\log\left|\dfrac{1-\sqrt{1+x^2}}{x} \right|+C$

Make use of symmetry, and integrate with respect to $y$ (rather than $x$).

$\dfrac{8\pi}{3}+4\sqrt{3}$

Use the symmetry of the function to re-write your integrals without an absolute value.

Area: $\dfrac{4}{3} - \sqrt[4]{\dfrac{4}{3}}$ Volume: $\dfrac{\pi^2}{6} - \dfrac{\sqrt{3}\pi}{4}$

Think of $e^x$ as $\left(e^{x/2}\right)^2$, and use a trig substitution. Then, use the identity $\sec^2 \theta = \tan^2 \theta +1$.

$2\sqrt{1+e^x}+2\log\left| 1-\sqrt{1+e^x} \right|-x+C$

Consider the ranges of the inverse trigonometric functions. For (c), also consider the domain of $\sqrt{x^2-a^2}$.

(a), (b): None. (c): $x< -a$