Definite integration when denominator consists of $x\sin x +1$

Question

$$\int_0^\pi\frac{x^2\cos^2x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx$$

The answer is $0$. I tried and made $(x\sin x +1)^2 $ in numerator and proceed, but not able to do any further.

Answer 1

First, let us try and simplify the integrand a bit. As \begin{align} \frac{x^2 \cos^2 x - x \sin x - \cos x - 1}{(1 + x \sin x)^2} &= \frac{x^2 - x^2 \sin^2 x - x \sin x - \cos x - 1}{(1 + x \sin x)^2}\\ &= \frac{x^2 + x \sin x - \cos x - (1 + x \sin x)^2}{(1 + x \sin x)^2}\\ &= -1 + \frac{x^2 + x \sin x - \cos x}{(1 + x\sin x)^2}, \end{align} we have for the integral $$I = -\pi + \int_0^\pi \frac{x^2 + x \sin x - \cos x}{(1 + x \sin x)^2} = -\pi + J.$$

To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).

Recall that if $u$ and $v$ are differentiable functions, from the quotient rule $$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$ and it is immediate that $$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$

For the integral $J$ we see that $v = 1 + x \sin x$. So $v' = \sin x + x \cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that $$u' v - v' u = u'(1 + x \sin x) - u (\sin x + x \cos x) = x^2 + x \sin x - \cos x.$$ After a little trial and error we find that if $$u = -x \cos x,$$ as $$u' = -\cos x + x \sin x,$$ this gives $$u' v - v' u = x^2 + x \sin x - \cos x,$$ as required.

Our integral can now be readily found as it can be rewritten in the form given by (1). The result is: \begin{align} I &= -\pi + \int_0^\pi \left (\frac{-x \cos x}{1 + x \sin x} \right )' \, dx\\ &= -\pi - \left [\frac{x \cos x}{1 + x \sin x} \right ]_0^\pi\\ &= -\pi + \pi\\ &= 0, \end{align} as announced.