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Questions  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Q)

Integrate : $\int \large\frac{\cos x - \sin x}{\sqrt {\sin 2x}}$$dx$

$(a)\;\log |(\sin x+\cos x ) +\sqrt {\cos 2x}|+c \\(b)\; \log |(\sin x+\cos x ) +\sqrt {\sin 2x}|+c \\(c)\;\sqrt {\sin 2x} \\ (d)\;None$

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A)
To solve this type of question we have to write
$\sin 2x =(\sin x +\cos x)^2 -1$
$\qquad= 1- (\sin x - \cos x)^2$
=> $\int \large\frac{\cos x - \sin x }{\sqrt {-1+ (\sin x +\cos x )^2 }}$$dx$
=> $(\sin x + \cos x )=t$
differentiate with respect to x
$(\cos x- \sin x )dx=dt$
=>$\int \large\frac{dt}{\sqrt {t^2-1}}$
=> $\log | t +\sqrt {t^2 -1}| +c$
$ \log |(\sin x+\cos x ) +\sqrt {\sin 2x}|+c$
Hence b is the correct answer.
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