Aspire Faculty ID #16428 · Topic: NIMCET 2009 · Just now
NIMCET 2009

The value of $\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x},dx$ is:

Solution

Let $I = \displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x}dx$ 
 Use property: $I = \int_{0}^{\pi} f(x)dx = \int_{0}^{\pi} f(\pi - x)dx$ 
 Compute 
$f(\pi - x) = \dfrac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} = \dfrac{(\pi - x)\sin x}{1 + \cos^2 x}$ 
 Add them: $f(x) + f(\pi - x) = \dfrac{\pi \sin x}{1 + \cos^2 x}$ 
 So, $2I = \displaystyle \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x}dx$ 
 Let $u = \cos x$, $du = -\sin x,dx$. 
When $x=0$, $u=1$, and when $x=\pi$, $u=-1$: 
$2I = \pi \displaystyle \int_{1}^{-1} \frac{-du}{1+u^2}$ 
$2I = \pi \displaystyle \int_{-1}^{1} \frac{du}{1+u^2}$ 
 This equals: $2I = \pi\left[\tan^{-1}u\right]_{-1}^{1} = \pi\left(\dfrac{\pi}{4} - \left(-\dfrac{\pi}{4}\right)\right)$ 
$2I = \pi \cdot \dfrac{\pi}{2} = \dfrac{\pi^2}{2}$ 
 So, $I = \dfrac{\pi^2}{4}$

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