trig integral

**mms** · October 7th 2009, 06:56 PM

prove that
$ \int\limits_0^{2\pi } {\frac{{xsen(x)}} {{1 + \cos ^2 (x)}}dx = } \frac{\pi } {2}\int\limits_{ - 1}^1 {\frac{1} {{1 + x^2 }}dx} $

and calculate the integral

**simplependulum** · October 7th 2009, 09:12 PM

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

Sub. $x = \pi - t$

$dx = -dt$

$I = \int_0^{\pi} \frac{(\pi-t)\sin(\pi - t)}{1 + \cos^2(\pi -t)}~dt$

$I = \int_0^{\pi} \pi \cdot \frac{\sin(t)}{1 + \cos^2(t)}~dt - I$

$\implies$

$2I = \pi \int_0^{\pi} \frac{\sin(t)}{1 + \cos^2(t)} ~dt$

Sub $\cos(t) = u$ $\sin(t) dt = -du$

$2I = \pi \int_{-1}^1 \frac{1}{1 + u^2}~du$

$I = \frac{\pi}{2} \int_{-1}^1 \frac{1}{1 + u^2}~du$

$I = (\frac{\pi}{2})^2$

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