Correct Answer is : −log⁡∣sin⁡x+cos⁡x+sin⁡2x∣+c- ≤ft| x+ x+{ 2 x}|+c−logsinx+cosx+sin2x+c

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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\int\frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x=

Solution:

Let

I=\int\frac{\sin x-\cos x}{\sqrt{\sin 2 x}} .d x

I=\int\frac{\sin x-\cos x}{\sqrt{1+2\sin x .\cos x-1}} d x

I=\int\frac{\sin x-\cos x}{\sqrt{(\cos x+\sin x)^2-1}} d x

Let

\sin x+\cos x=t

\Rightarrow \cos x-\sin x=\frac{dt}{dx}

\Rightarrow -(\sin x-\cos x) d x = d t

\Rightarrow (\sin x-\cos x) d x = -d t

\therefore I=-\int\frac{dt}{\sqrt{t^2-1}}

\left[\because \int\frac{1}{\sqrt{x^2-a^2}} d x = \log \left|x+\sqrt{x^2-a^2}\right|+C \right]

=-\log \left|t+\sqrt{t^2-1}\right|

=-\log \left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^2}-1\right|+C

I=-\log \left|\sin x+\cos x+\sqrt{\sin 2 x}\right|+C

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