`int_(pi/2)^pi sinx/(1+cos^2x) dx` Evaluate the definite integral

Monday, July 4, 2011

We have to evaluate the integral: $\int_{\pi/2}^{\pi}\frac{sinx}{1+cos^2x}dx$

Let $\displaystyle{\cos{{x}}}={u}$

So, $\displaystyle-{\sin{{x}}}{\left.{d}{x}\right.}={d}{u}$

When $x=\pi/2, u=0$

$x=\pi, u=-1$

So we have,

$\int_{\pi/2}^{\pi}\frac{sinx}{1+cos^2x}dx=\int_{0}^{-1}\frac{-du}{u^2+1}$

$=\int_{-1}^{0}\frac{du}{u^2+1}$

=arctan(0)-arctan(-1)

=-arctan(-1)

=arctan(1)

= $\displaystyle\frac{\pi}{{4}}$

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