Evaluate int 0 to pi x cos x upon 1 plus cos square x

Question 2:

Evaluate: $\displaystyle \int_0^{\pi} \frac{x\sin x}{1 + \cos^2x}dx$

Let $I= \displaystyle \int_0^{\pi} \frac{x\sin x}{1 + \cos^2x}dx$ ………….(i)

Using property

$\int_0^a f(x) dx = \int_0^af(a - x) dx$

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

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

$\Rightarrow I = \displaystyle \int_0^{\pi} \frac{\pi\sin x}{1 + \cos^2x}dx- \int_0^{\pi} \frac{x\sin x}{1 + \cos^2x}dx$

$\Rightarrow I = \displaystyle \int_0^{\pi} \frac{\pi\sin x}{1 + \cos^2x}dx- I$

$\Rightarrow 2I = \pi \displaystyle \int_0^{\pi} \frac{\sin x}{1 + \cos^2x}dx$ ……….(ii)

Put $\cos x = t$

$\Rightarrow -\sin x = \frac{dt}{dx}$

$\Rightarrow dx = \frac{dt}{-\sin x}$

New limits, when x = 0, t ⇒ 1 and x = π ⇒ t = -1

Putting in eq (ii)

$\Rightarrow 2I = \pi \displaystyle \int_1^{-1} \frac{\sin x}{1 + (t)^2}\frac{dt}{-\sin x}$

$\Rightarrow 2I = -\pi \displaystyle \int_1^{-1} \frac{dt}{1 + (t)^2}$

$\Rightarrow 2I = -\pi \displaystyle [\tan^{-1}t]_1^{-1}$

$\Rightarrow 2I = -\pi \displaystyle[\tan^{-1}(-1) -\tan^{-1}(1)]$

$\Rightarrow 2I = -\pi \displaystyle[\frac{-\pi}{4} - \frac{\pi}{4}]$

$\Rightarrow 2I = -\pi [-\frac{2\pi}{4}]$

$\Rightarrow 2I = \dfrac{\pi^2}{2}$

$\Rightarrow I = \dfrac{\pi^2}{4}$