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

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  • $\int\limits_a^b f(x)dx=F(b)-F(a)$
  • $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Step 1:
$I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\sin x-\cos x}{1+\sin x\cos x}$$dx-----(1)$
Applying the property $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\sin (\pi/2-x)-\cos (\pi/2-x)}{1+\sin (\pi/2-x)\cos (\pi/2-x)}$$dx$
But $\sin (\pi/2-x)=\cos x\;and\;\cos(\pi/2-x)=\sin x$
Therefore $I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\cos x-\sin x}{1+\cos x\sin x}$$dx-----(2)$
Step 2:
Adding equ(1) and equ(2)
$2I=\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\sin x-\cos x}{1+\sin x\cos x}+\int\limits_0^\frac{\Large \pi}{\Large 2}\large\frac{\cos x-\sin x}{1+\cos x\sin x}$
On simplifying we get,
answered Oct 1, 2013 by sreemathi.v
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