I was trying to solve the following problem from MIT 2006 Integration Bee: $$
\int_{0}^\infty \frac{dx}{\left(x+\sqrt{1+x^2}\right)^2}.
$$ I've managed to reduce this integral to $$
\int_{0}^{\pi/2} \frac{d\theta}{(\sin\theta+1)^2}
$$ using the trigonometric substitution $x=\tan\theta$. From here I don't know how to proceed. I've consulted symbolab.com. It gave me the answer $2/3$, which is consistent with the answer provided by Integration Bee. In its solution there was this step $$
\int_{0}^{\pi/2} \frac{d\theta}{(\sin\theta+1)^2} = \int_0^1 \frac{2(u^2+1)}{(u^2+2u+1)^2}du.
$$ I couldn't figure out what substitution it used. Any ideas? Thanks for reading my post.
Best Answer
1. Computing the original integral. I guess this is the shortest way to solve the integral: Substituting $x = \sinh t$,
$$ \int_{0}^{\infty} \frac{\mathrm{d}x}{(x+\sqrt{1+x^2})^2} = \int_{0}^{\infty} \frac{\cosh t \, \mathrm{d}t}{\left(\sinh t + \cosh t\right)^2} = \int_{0}^{\infty} \frac{e^{-t} + e^{-3t}}{2} \, \mathrm{d}t = \frac{2}{3}. $$
2. What substitution has been used? Note the Weierstrass substitution:
$$ u = \tan(\theta/2), \qquad \mathrm{d}\theta = \frac{2\,\mathrm{d}u}{1+u^2}, \qquad \sin\theta = \frac{2u}{1+u^2}, \qquad \cos\theta = \frac{1-u^2}{1+u^2}. $$
Then as $\theta$ runs from $0$ to $\frac{\pi}{2}$, $u$ runs from $0$ to $1$. Hence,
$$ \int_{0}^{\pi/2} \frac{\mathrm{d}\theta}{(\sin\theta+1)^2} = \int_{0}^{1} \frac{1}{(\frac{2u}{1+u^2} + 1)^2} \cdot \frac{2\,\mathrm{d}u}{1+u^2}. $$
Simplifying the expression gives the desired rational integral.