JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c\) where \(c\) is a constant of integration, then the ordered pair \(( a , b )\) is equal to
- A \((-1,3)\)
- B \((3,1)\)
- C \((1,3)\)
- D \((1,-3)\)
Answer & Solution
Correct Answer
(C) \((1,3)\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x\) \(=\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x)^{2}}} d x\) Let \(\sin x+\cos x=t\) \(\int \frac{d t}{\sqrt{9-t^{2}}}=\sin ^{-1} \frac{t}{3}+c\) \(=\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c\) So \(a=1, b=3\)
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