Find the integral of the function x- x 1+ 2 x - Vidyadip

Question

Find the integral of the function $\frac{\cos x-\sin x}{1+\sin 2 x}$

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Answer

Clearly,
$\frac{\cos x-\sin x}{1+\sin 2 x}$ $= \frac{\cos x-\sin x}{\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cos x}$
$= \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}}$
Let sinx + cosx = t
$\Rightarrow$ (cosx-sinx)dx = dt
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$
$= \int \frac{d t}{t^{2}}$
= -t^-1 + C
$=-\frac{1}{t}+C$
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x = \frac{-1}{\sin x+\cos x}+C$

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