MHT CET · Maths · Indefinite Integration
If \(\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} \mathrm{~d} x=\operatorname{asin}^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c\)
Where c is a constant of integration, then the ordered pair \((\mathrm{a}, \mathrm{b})\) is equal to
- A \((1,3)\)
- B \((3,1)\)
- C \((-1,3)\)
- D \((-3,1)\)
Answer & Solution
Correct Answer
(A) \((1,3)\)
Step-by-step Solution
Detailed explanation
\(\text {Let } \mathrm{I} =\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x \)
\( =\int \frac{\cos x-\sin x}{\sqrt{9-(1+\sin 2 x)}} d x \)
\( =\int \frac{\cos x-\sin x}{\sqrt{3^2-(\cos x+\sin x)^2}} \mathrm{~d} x\)
Put \(\cos x+\sin x=\mathrm{t}\)
\(\Rightarrow(-\sin x+\cos x) \mathrm{d} x=\mathrm{dt}\)
\(\therefore I =\int \frac{d t}{\sqrt{3^2-t^2}} d t \)
\( =\sin ^{-1}\left(\frac{t}{3}\right)+c \)
\( =\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c \)
\( \therefore (a, b)=(1,3)\)
\( =\int \frac{\cos x-\sin x}{\sqrt{9-(1+\sin 2 x)}} d x \)
\( =\int \frac{\cos x-\sin x}{\sqrt{3^2-(\cos x+\sin x)^2}} \mathrm{~d} x\)
Put \(\cos x+\sin x=\mathrm{t}\)
\(\Rightarrow(-\sin x+\cos x) \mathrm{d} x=\mathrm{dt}\)
\(\therefore I =\int \frac{d t}{\sqrt{3^2-t^2}} d t \)
\( =\sin ^{-1}\left(\frac{t}{3}\right)+c \)
\( =\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c \)
\( \therefore (a, b)=(1,3)\)
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