Question

Evaluate:

$\displaystyle \sf \int_0^\pi \dfrac{x \ sin \ x}{1 + cos^2 \ x} \ dx$

Answer 1

Solution :

Before we solve the problem, let's get to know about some properties as follows :

1. $\displaystyle \int_{0}^{a}f(x)\:dx=\int_{0}^{a}f(a-x)\:dx$

2. $\displaystyle \int_{a}^{b}f(x)\:dx=-\int_{b}^{a}f(x)\:dx$

Now, we proceed to solve the given integration problem.

Let, I = $\displaystyle \int_{0}^{\pi}\frac{x\:sinx}{1+cos^{2}x}dx$

= $\displaystyle \int_{0}^{\pi}\frac{(\pi-x)\:sin(\pi-x)}{1+cos^{2}(\pi-x)}dx$

= $\displaystyle \small \int_{0}^{\pi}\frac{\pi\:sinx}{1+cos^{2}x}dx-\int_{0}^{\pi}\frac{x\:sinx}{1+cos^{2}x}dx$

or, I = $\displaystyle \pi \int_{0}^{\pi}\frac{sinx}{1+cos^{2}x}dx$ - I

or, 2I = $\displaystyle \pi \int_{0}^{\pi}\frac{sinx\:dx}{1+cos^{2}x}$

Let, cosx = z so that

- sinx dx = dz

and as x limits from 0 to π, z limits from 1 to - 1

Thus, 2I = $\displaystyle -\pi \int_{1}^{-1}\frac{dz}{1+z^{2}}$

= $\displaystyle \pi \int_{-1}^{1}\frac{dz}{1+z^{2}}$

= $\displaystyle \pi [tan^{-1}z]_{-1}^{1}$

= π {π/4 - (- π/4)}

or, 2I = π (π/2)

or, I = π²/4

Therefore, $\displaystyle \int_{0}^{\pi}\frac{x\:sinx}{1+cos^{2}x}dx=\frac{{\pi}^{2}}{4}$

Answer 2

Answer:

Step-by-step explanation: