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The value of $\int\limits_{0}^{\pi/2}\large\frac{\sin x-\cos x}{1+\sin x \cos x}$$dx$ is

\[\begin{array}{1 1}(1)\frac{\pi}{2}&(2)0\\(3)\frac{\pi}{4}&(4)\pi\end{array}\]

1 Answer

$I=\int\limits_{0}^{\pi/2}\large\frac{\sin x-\cos x}{1+\sin x \cos x}$$dx$ -----(1)
$I=\int\limits_{0}^{\pi/2}\large\frac{\sin (\frac{\pi}{2} -x)- \cos (\frac{\pi}{2}-x)}{1+\sin (\frac{\pi}{2}-x) \cos (\frac{\pi}{2} -x)}$$dx$
$I=\int\limits_{0}^{\pi/2}\large\frac{\cos x-\sin x}{1+ \cos x \sin x}.$$dx$ -----(2)
Equation (1)+(2) => $2I=\int\limits_{0}^{\pi/2} \large\frac{\cos x-\sin x}{1+\sin x \cos x}$$dx+\int\limits_{0}^{\pi/2} \large\frac{\cos x-\sin x}{1+\sin x \cos x}$$dx$
$\qquad= 0$
hence 2 is the correct answer.
answered May 20, 2014 by meena.p
 
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