Hairy $u$-substitution problem

Question

Evaluate using the $u$-substitution:

$$\int_{-1}^{1} \frac{dx}{4 + x^2}.$$

Now, I was told to set $$\tan u = \frac{x}{2},$$ but that doesn't help me at all. Hints needed!

Answer 1

Well, the substitution $\tan u = x/2$ is unnecessary because $$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$$ is a well-known result. In your case $a = 2$.

However, if you insist, using the substitution $\tan u = x/a$, we can actually derive the formula above. To this end, we proceed as follows.

Say $x = a\tan u$. Then $dx/du = a \sec^2 u$ and $u = \tan^{-1}(x/a)$, so $$\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 u}{a^2(1 + \tan^2 u)} du = \int \frac{du}{a} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C.$$ Notice that I used the identity $1 + \tan^2 u = \sec^2 u.$