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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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  • (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,


answered Feb 12, 2013 by meena.p
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