# Trigonometric Substitutions

## Integrals containing $$\sqrt{a^2-x^2}$$

One of the substitutions $$x=a\cos\theta$$ and $$x=a\sin \theta$$ may simplify the integral.

Example 1

Evaluate the integral $$\displaystyle \int_0^a \frac1{\sqrt{a^2-x^2}} \,dx$$.

Example 2

Evaluate the integral $$\displaystyle \int_0^a \sqrt{a^2-x^2} \,dx$$.

## Integrals containing $$\sqrt{x^2+a^2}$$

The substitution $$x=a\tan\theta$$ has $$dx=\frac{a}{\cos^2\theta}\,d\theta$$ and has a convenient property that $$a^2+x^2=a^2\left(1+\tan^2\theta\right)=\frac{a^2}{\cos^2\theta}$$.

Example 3

Evaluate the integral $$\displaystyle \int \sqrt{x^2+a^2} \,dx$$.

There is another substitution that can be used to simplify integrals containing $$\sqrt{x^2+a^2}$$. We can use $$x=a\sinh \theta$$, where $$\displaystyle \sinh u=\frac{e^u-e^{-u}}2$$. This substitution has $$dx=a\cosh \theta$$ and $$x^2+a^2=a^2\left(\sinh^2\theta+1\right)=a^2\cosh^2\theta$$. We will illustrate how to use hyperbolic substitutions in the next section.

## Integrals containing $$\sqrt{x^2-a^2}$$

There are two ways to evaluate integrals involving this type of square roots. One is the substitution $$x=a\sec \theta$$. The other way is using substitution $$x=a\coth \theta$$. Substitution with secant is very similar to the one with tangent we used in the previous problem. Let us illustrate the substitution with hyperbolic cotangent.

We will use the substitution $$\displaystyle x=a\coth\theta=\frac{\sinh \theta}{\cosh\theta}$$. Recall that $$\displaystyle \cosh \theta=\frac{e^{\theta}+e^{-\theta}}2$$ and $$\displaystyle \sinh\theta=\frac{e^{\theta}-e^{-\theta}}2$$. It is easy to verify that $$\cosh^{\prime}\theta=\sinh\theta$$, $$\sinh^{\prime}\theta=\cosh\theta$$, $$\cosh^2\theta-\sinh^2\theta=1$$, $$\displaystyle \coth^{\prime}\theta=\frac1{\sinh^2\theta}$$. Therefore $$\displaystyle dx=\frac{a}{\sinh^2\theta}\,d\theta$$ and $$\displaystyle x^2-a^2=a^2\left(\coth^2\theta-1\right)=\frac{a^2}{\sinh^2\theta}$$.

Example 4

Evaluate the integral $$\displaystyle \int \sqrt{x^2-a^2} \,dx$$.

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