MHT CET · Maths · Indefinite Integration
\(\int \frac{\mathrm{d} x}{\cot ^2 x-1}=\frac{1}{\mathrm{~A}} \log |\sec 2 x+\tan 2 x|-\frac{x}{\mathrm{~B}}+\mathrm{c}\), (where \(\mathrm{c}\) is constant of integration), then \(\mathrm{A}+\mathrm{B}=\)
- A \(-6\)
- B \(6\)
- C \(-5\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\mathrm{d} x}{\cot ^2 x-1} =\int \frac{\mathrm{d} x}{\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x}} \)
\( =\int \frac{\sin ^2 x}{\cos 2 x} \mathrm{~d} x \)
\( =\int \frac{\frac{1-\cos 2 x}{2}}{\cos 2 x} \mathrm{~d} x \)
\( =\frac{1}{2} \int(\sec 2 x-1) \mathrm{d} x \)
\( =\frac{1}{2}\left(\frac{\log |\sec 2 x+\tan 2 x|}{2}-x\right)+\mathrm{c} \)
\( =\frac{1}{4} \log |\sec 2 x+\tan 2 x|-\frac{x}{2}+\mathrm{c} \)
\( \therefore \mathrm{A}=4, \mathrm{~B} =2 \)
\( \Rightarrow \mathrm{A}+\mathrm{B} =6\)
\( =\int \frac{\sin ^2 x}{\cos 2 x} \mathrm{~d} x \)
\( =\int \frac{\frac{1-\cos 2 x}{2}}{\cos 2 x} \mathrm{~d} x \)
\( =\frac{1}{2} \int(\sec 2 x-1) \mathrm{d} x \)
\( =\frac{1}{2}\left(\frac{\log |\sec 2 x+\tan 2 x|}{2}-x\right)+\mathrm{c} \)
\( =\frac{1}{4} \log |\sec 2 x+\tan 2 x|-\frac{x}{2}+\mathrm{c} \)
\( \therefore \mathrm{A}=4, \mathrm{~B} =2 \)
\( \Rightarrow \mathrm{A}+\mathrm{B} =6\)
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