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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