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# Evaluate the integral using substitution$\int\limits_0^\frac{\large \pi}{2}\frac{\sin x}{1+\cos^2x}dx$

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Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• Method of substitution:If $I=\int f(x)dx,Let f(x)=t\; then\; f'(x)dx=dt\; therefore\; I=\int t.dt$
• (iii)$\int \large \frac{\sin x}{1+x^2}dx=\tan^{-1}(x)$
Given $\int\limits_0^\frac{\large \pi}{2}\frac{\sin x}{1+\cos^2x}dx$

Let $\cos x=t,$ on differentiating with respect to x

$-\sin xdx=dt\qquad=>\sin x dx=-dt$

Now substituting t and dt,

$I=\int \limits_0^1 \large\frac{dt}{1+t^2} \;As\; x \to \pi/2,t \to 0$

$=-\int \limits_0^1\frac {dt}{1+t^2}$

This is of the form $=\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan ^{-1}(\frac{x}{a})$

Here $a=1\;and\;x=t$

Therefore $\int \limits_0^1\large\frac{dt}{1+t^2}=\tan^{-1}(t)$

On applying limits we get,

$=\tan^{-1}(1)$

$I=\large\frac{\pi}{4}$

answered Feb 12, 2013 by