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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals2 Marks
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 $\sin x + \cos x = t$
$\Rightarrow (\cos x-\sin x)dx = dt$
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$
$= \int \frac{d t}{t^{2}}$
$= -t^{-1} + C$
$=-\frac{1}{t}+C$
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d\ x $
$= \frac{-1}{\sin x+\cos x}+C$

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