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Prove that $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ , hence evaluate $\int_0^\pi \frac{x\sin x}{1 + cos^2 x}dx$ .

Answers (1)

To prove: $\int_0^a f(x)dx = \int_0^a f(a-x)dx$

Let $x = a -t$

$dx = -dt$

$\\x = 0, t = a \\x = a, t = 0$

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

$\int_0^a f(x)dx = -\int^0_a f(a-t)dt$

[ taking minus sign to invert the integral $\int_b^a f(x)dx = -\int^b_a f(x)dx$ ]

$\Rightarrow \int_0^a f(x)dx = \int^a_0 f(a-x)dx$

Hence Proved

$\int_0^\pi \frac{x\sin x}{1 + cos^2 x}dx$ .

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

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

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

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

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

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

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

$\\\cos x = t \\-\sin x = dt \\ \sin x = - dt \\ x\rightarrow 0 \quad t = 1 \\ x = \pi \quad t = -1$

$2I =\pi\int_{1}^{-1} \frac{-dt}{1 + t^2}$

$I =\frac{1}{2}\pi\int^{1}_{-1} \frac{dt}{1 + t^2}$

$I =\frac{\pi}{2}\int^{1}_{-1} \frac{dt}{1 + t^2}$

$I =\frac{\pi}{2}\left[\tan^{-1}t \right ]_{t=-1}^{t = 1}$

$I =\frac{\pi}{2}\left[\tan^{-1}1 - \tan^{-1}(-1) \right ]$

$I =\frac{\pi}{2}\left[\frac{\pi}{4} - \left(-\frac{\pi}{4} \right ) \right ]$

$= \frac{\pi^2}{4}$

Posted by

Ravindra Pindel

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