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# 2.2: Trigonometric Substitution

When we have integrals that involve the square root term

$\sqrt{a^2+x^2}$

we may be able to trigonometric substitution to solve the integral.

Example 1

Solve

$\int \sqrt{1-x^2}dx$

by substituting $$x=\sin \theta$$ and $$dx=\cos \theta \, d\theta$$.

The integrand then becomes

\begin{align} \sqrt{1-x^2} &= \sqrt{1-\sin^2 \theta} \\ &= \sqrt{\cos^2 \theta} \\ &= \cos \theta \end{align}

We have

\begin{align} \int \sqrt{1-x^2}\; dx &= \int \cos\theta\cos\theta \;d\theta \\ &= \int \cos^2\theta \;d\theta \\ &= \int \Big( \dfrac{1}{2}+\dfrac{1}{2}\cos 2\theta \Big)\; d\theta \\ &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta +C \\ &= \dfrac{1}{2} \arcsin x +\dfrac{1}{2} \sin\theta\cos\theta+C \\ &= \dfrac{1}{2}\arcsin x +\dfrac{1}{2}x\sqrt{1-x^2}+C \end{align}

Exercises

1. $\int \dfrac{\sqrt{1-x^2}}{x^4} dx$

2. $\int \dfrac{1}{\sqrt{4-9x^2}}dx$

### Two Key Formulas

From Trigonometry, we have the following two key formulas:

$\sec^2\, x = 1 + \tan^2\, x$

so

$\sec x = \sqrt{1+\tan^2 x}$

and

$\tan^2\, x = \sec^2\, x - 1$

so

$\tan x = \sqrt{\sec^2 \, x -1}.$

When we have integrals that involve any of the above square roots, we can use the appropriate substitution.

Example 2

\begin{align} \int \dfrac{x^3}{\sqrt{1+x^2}}dx \\ x= \tan\theta, \; dx=\sec^2\theta \; d\theta \\ \sqrt{1+x^2}=\sqrt{1+\tan^2\theta}= \sqrt{\sec^2\theta} = \sec\theta \\ &= \int \dfrac{\tan^3\theta}{\sec\theta}\sec^2\theta \; d\theta \\ &= \int \tan^3\theta \sec\theta \; d\theta \\ &= \int \tan^2\theta \tan\theta \sec\theta \; d\theta \\ &= \int (\sec^2\theta-1)\sec\theta\tan\theta \; d\theta \\ u=\sec\theta, \; du=\sec\theta\tan\theta \; d\theta \\ &= \int (u^2-1) \; du \\ &=\dfrac{u^3}{3}-u+C \\ &= \dfrac{\sec^3\theta}{3}-\sec\theta+C \\ &= \dfrac{(1+x^2)^{\frac{3}{2}}}{3}-\sqrt{1+x^2}+C \end{align}

Exercise

1. $\int \dfrac{x^3}{\sqrt{x^2-1}} \; dx$

2. $\int \dfrac{x^2}{\sqrt{9+4x^2}} \; dx$

3. $\int \dfrac{1}{\sqrt{x^2+2x}} \; dx$

### Contributors

• Integrated by Justin Marshall.