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If \int\frac{\cos x-\sin x}{\sqrt{8-\sin2x}}dx=a\sin ^{-1}\left ( \frac{\sin x+\cos x}{b} \right )+c, where c is a constant of integration, then the ordered pair (a,b) is equal to :
Option: 1 (1,-3)
Option: 2 (3,1)
Option: 3 (-1,3)
Option: 4 (1,3)

Answers (1)

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\text { put } \sin x+\cos x=t \Rightarrow 1+\sin 2x=\mathrm{t}^{2}

\Rightarrow(\cos x-\sin x) \mathrm{d} x=\mathrm{dt}

\\\therefore \quad I=\int \frac{d t}{\sqrt{8-\left(t^{2}-1\right)}}=\int \frac{d t}{\sqrt{9-t^{2}}}\\\\=\sin ^{-1}\left(\frac{t}{3}\right)+C=\sin ^{-1}\left(\frac{\sin \theta+\cos \theta}{3}\right)+C \\\\ \Rightarrow a=1 \text { and } b=3

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Suraj Bhandari

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