7. Trigonometric Substitutions

a1. Substitutions for $x^2+1$ - Tangent Substitutions

Recall the derivative and integral formulas for $\arctan x$: $\dfrac{d}{dx}\arctan x=\dfrac{1}{x^2+1} \qquad \qquad \qquad \int \dfrac{1}{x^2+1}\,dx=\arctan x+C$ and the second Pythagorean Trig Identity: $\tan^2\theta+1=\sec^2\theta$

These formulas motivate the following integration technique:

Here are some examples:

Compute $\displaystyle \int \dfrac{1}{\sqrt{x^2+1}}\,dx$.

We substitute $x=\tan\theta$. Then $dx=\sec^2\theta\,d\theta$ and so $\begin{aligned} \int \dfrac{1}{\sqrt{x^2+1}}\,dx &=\int \dfrac{1}{\sqrt{\tan^2\theta+1}}\,\sec^2\theta\,d\theta \\ &=\int \dfrac{1}{\sqrt{\sec^2\theta}}\,\sec^2\theta\,d\theta =\int \sec\theta\,d\theta \\ &=\ln|\sec\theta+\tan\theta|+C \end{aligned}$ Don't forget to substitute for the differential! Don't forget to substitute back! We already know that $\tan\theta=x$ from the definition of the substitution. We also need a formula for $\sec\theta$ in terms of $x$. There are two ways to find this formula:

We can use the Pythagorean identity $\tan^2\theta+1=\sec^2\theta$. Then: $\sec\theta=\sqrt{\tan^2\theta+1}=\sqrt{x^2+1}$

Returning to the problem at hand, we conclude: $\int \dfrac{1}{\sqrt{x^2+1}}\,dx =\ln\left|\sqrt{x^2+1}+x\right|+C$

We check by differentiating. If $f(x)=\ln\left|\sqrt{x^2+1}+x\right|$ then $\begin{aligned} f'(x) &=\dfrac{1}{\sqrt{x^2+1}+x}\left(\dfrac{1}{2}\dfrac{2x}{\sqrt{x^2+1}}+1\right) \\ &=\dfrac{1}{\sqrt{x^2+1}+x}\left(\dfrac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\right) =\dfrac{1}{\sqrt{x^2+1}} \end{aligned}$ which is the integrand we started with.

Compute $\displaystyle \int \dfrac{x}{\sqrt{x^2+1}}\,dx$.

Method 1:
We substitute $x=\tan\theta$. Then $dx=\sec^2\theta\,d\theta$ and so: $\begin{aligned} \int \dfrac{x}{\sqrt{x^2+1}}\,dx &=\int \dfrac{\tan\theta}{\sqrt{\tan^2\theta+1}}\sec^2\theta\,d\theta =\int \dfrac{\tan\theta}{\sqrt{\sec^2\theta}}\sec^2\theta\,d\theta \\ &=\int \tan\theta\sec\theta\,d\theta=\sec\theta+C =\sqrt{x^2+1}+C \end{aligned}$ This solution was unnecessarily complicated. Always look for an ordinary substitution first:

Method 2:
Let $u=x^2+1$. Then $du=2x\,dx$ or $\dfrac{1}{2}\,du=x\,dx$. So: $\int \dfrac{x}{\sqrt{x^2+1}}\,dx =\dfrac{1}{2}\int \dfrac{1}{\sqrt{u}}\,du =\sqrt{u}+C=\sqrt{x^2+1}+C$

Now it's your turn: