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JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 70

If ∫cos⁡x−sin⁡x8−sin⁡2xdx=asin⁡−1(sin⁡x+cos⁡xb)+c\displaystyle\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 :

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Answer and worked solution

Official answer: B

Substitute t=sin⁡x+cos⁡xt=\sin x+\cos x: 8−sin⁡2x=9−t28-\sin2x=9-t^2 and dt=(cos⁡x−sin⁡x)dxdt=(\cos x-\sin x)dx, giving sin⁡−1t3\sin^{-1}\frac t3. So (a,b)=(1,3)(a,b)=(1,3).

(B)

Topics:Substitution
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