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) $x=5\tan\theta$

In the text, there is a template for choosing an appropriate substitution, but for this problem we will explain the logic of the choices.

The trig identities that we can use are:

They have the following forms:

In order to cancel out the square root, we should choose a substitution that will match the argument under the square root with the trig identity of the corresponding form.

(a) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function $9x^2-16$, which has the form (function) $-$ (constant). This form matches the trig identity $\sec^2 \theta - 1 = \tan^2 \theta$. We can set $x$ to be whatever we need it to be, but we don't have the same control over the constant, 16. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 16, we get

What we want is a substitution that gives us

Using this substitution,

So, we eliminated the square root.

(b) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function $1-4x^2$, which has the form (constant) $-$ (function). This form matches the trig identity $1-\sin^2\theta = \cos^2\theta$. What we want is a substitution that gives us

Using this substitution,

So, we eliminated the square root. We remark that the absolute value signs are not needed in $|\cos\theta|$, because, for $-\frac{1}{2}\le x\le\frac{1}{2}$, we have $\theta=\arcsin(2x)$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and $\cos(\theta)\ge 0$ for those $\theta$'s.

(c) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the fractional power. (That is, we want to eliminate the square root.) The function under the power is $25+x^2$, which has the form (constant) $+$ (function). This form matches the trig identity $\tan^2 \theta + 1 = \sec^2 \theta$. We can set $x$ to be whatever we need it to be, but we don't have the same control over the constant, 25. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 25, we get

What we want is a substitution that gives us

Using this substitution,

So, we eliminated the square root. We remark that the absolute value signs are not needed in $|\sec\theta|$, because, for $-\infty< x<\infty$, we have $\theta=\arctan(x/5)$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and $\sec(\theta)\ge 0$ for those $\theta$'s.

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

Just as in Question 1, we want to choose a trigonometric substitution that will allow us to eliminate the square roots. Before we can make that choice, though, we need to complete the square. In subsequent problems, we won't show the algebra behind completing the square, but for this problem we'll work it out explicitly. After some practice, you'll be able to do this step in your head for many cases.

After the squares are completed, the choice of trig substitution follows the logic outlined in the solutions to Question 1, or (equivalently) the template in the text.

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.

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

Let $x = 2\tan\theta$, so that $x^2+4 = 4\tan^2\theta+4=4\sec^2\theta$ and $\dee{x} = 2\sec^2\theta\,\dee{\theta}$. Then

To find $\sin\theta$ in terms of $x$, we construct the right triangle above. Since $\tan\theta = \dfrac{x}{2} = \dfrac{\text{opp}}{\text{adj}}$, we label the opposite side $x$ and the adjacent side $2$. By the Pythagorean Theorem, the hypotenuse has length $\sqrt{x^2+4}$. Then $\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{x}{\sqrt{x^2+4}}$.

To see why we could write $(\sec^2\theta)^{3/2} =\sec^3\theta$, as opposed to $(\sec^2\theta)^{3/2} =\big|\sec^3\theta\big|$, in the second line above, see Example 1.9.5 in the CLP-2 text.

As a check, we observe that the derivative of the answer

is exactly the integrand.

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.

$\dfrac{\pi}{6}$

Make the change of variables $x=5\sin\theta$, $\dee{x}=5\cos\theta\,\dee{\theta}$. Since $x=0$ corresponds to $\theta=0$ and $x=\frac{5}{2}$ correponds to $\sin\theta=\half$ or $\theta =\frac{\pi}{6}$,

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.

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

Substitute $x=5\tan u$, so that $\dee{x}=5 \sec^2u\,\dee{u}$.

To find $\sec u$ and $\tan u$, we have two options. One is to set up a right triangle with angle $u$ and $\tan u = \frac{x}{5}$. Then we can label the opposite side $x$ and the adjacent side 5, and use Pythagoras to find that the hypotenuse is $\sqrt{x^2+25}$.

Another option is to look back at our work a little more closely–in fact, we've already found what we're looking for. Since we used the substitution $x=5\tan u$, this gives us $\tan u = \frac{x}{5}$. In the denominator of the integrand, we simplified $\sqrt{x^2+25} = 5\sec u$, so $\sec u = \frac{1}{5}\sqrt{x^2+25} = \sqrt{1+\frac{x^2}{25}}$.

To see why we could write $\sqrt{x^2+25} =5\sec u$, as opposed to $\sqrt{x^2+25} =5|\sec u|$, see Example 1.9.5 in the CLP-2 text.

A trig substitution is not the easiest path.

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

The quadratic formula underneath the square root makes us think of a trig substitution, but in the interest of developing good habits, let's check for an easier way first. If we let $u=2x^2+4x$, then $\dee{u} = (4x+4)~\dee{x}$, so $\frac{1}{4}\,\dee{u}=(x+1)\,\dee{x}$. This substitution looks easier than a trig substitution (which would start with completing the square).

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

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

Substitute $x=4\tan u$, $\dee{x}=4 \sec^2u\,\dee{u}$.

To finish off the integral, we'll substitute $v=\sin u$, $\dee{v}=\cos u\,\dee{u}$.

To find $\sin u$, we draw a right triangle with angle $u$ and $\tan u = \frac{x}{4}$. We label the opposite side $x$ and the adjacent side $4$, and then from Pythagoras we find that the hypotenuse has length $\sqrt{x^2+16}$. So, $\sin u = \dfrac{\sqrt{x^2+16}}{x}$.

As a check, we observe that the derivative of the answer

is exactly the integrand.

The integrand should simplify quite far after your substitution.

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

Substitute $x=3\sec u$ with $0\le u<\frac{\pi}{2}$. Then $\dee{x}= 3\sec u\tan u\,\dee{u}$ and $\sqrt{x^2-9}=\sqrt{9\sec^2 u-9} =\sqrt{ 9\tan^2 u}=3\tan u$, so that

To evaluate $\sin u$, we make a right triangle with angle $u$. Since $\sec u = \dfrac{x}{3} = \dfrac{\text{hyp}}{\text{adj}}$, we label the hypotenuse $x$ and the adjacent side $3$. Using the Pythagorean Theorem, the opposite side has length $\sqrt{x^2-9}$. So, $\sin u = \dfrac{\sqrt{x^2-9}}x$ and

As a check, we observe that the derivative of the answer

is exactly the integrand. (We remark that this is the case even for $x\le -3$.)

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,