A calculus problem by Kishore S. Shenoy

Relevant wiki: Definite Integrals

A substitution $cosx=t$ yields $\displaystyle J=\int_{-1}^{1} \frac{cos^{-1}x}{1+x^2}dx$

Since $\displaystyle cos^{-1}(-x) = \pi-cos^{-1}x$ is valid when $x\in(-1,1)$ so we have ,

$\displaystyle J = 2\int_{0}^{1}\frac{\pi}{1+x^2}dx - J\implies J =\pi[tan^{-1}x]_{0}^{1} = \frac{\pi^{\color{#D61F06}{2}}}{\color{#20A900}{4}}$

Thus the answer $\boxed{2+4+1=7}$

Relevant wiki: Definite Integrals

Let $f(x) = \dfrac{x}{2-x^2}\\\Rightarrow I = \int_0^\pi xf(\sin x)\, dx$

Now using properties of definite integral, $\displaystyle\int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx$ $\begin{aligned}I &= \int_0^\pi xf(\sin x)\, dx\\ &= \int_0^\pi (\pi-x)f(\sin (\pi-x))\, dx\\&=\int_0^\pi (\pi-x)f(\sin x)\, dx\end{aligned}$

Adding both the equations, $I = \dfrac\pi2\int_0^\pi f(\sin x)\, dx$

Now using the property $\displaystyle\int_0^{2a}f(x)\,dx = \int_0^a f(x) + f(2a-x)\,dx$ , we get $\begin{aligned}I &= \pi \int_0^{\pi/2} \dfrac{\sin x}{1+\cos^2x}\, dx \quad\color{#3D99F6}{t = \cos x,~dt = -\sin x\, dx}\\&=\int_0^1\dfrac 1{1+t^2}\, dt\\&=\pi\left(\dfrac\pi 4 - 0\right)\\&=\large\boxed{\dfrac{\pi^2}{4}}\end{aligned}$

Note.

We can say generally that $\boxed{\displaystyle\int_0^\pi x f(\sin x)\, dx = \pi\int_0^{\pi/2}f(\sin x)\, dx}$