JEE Mains · Maths · STD 12 - 7.1 indefinite integral
Let \(I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x\). If \(I(0)=3\), then \(\mathrm{I}\left(\frac{\pi}{12}\right)\) is equal to :
- A \(\sqrt{3}\)
- B \(3 \sqrt{3}\)
- C \(6 \sqrt{3}\)
- D \(2 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(3 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\( I(x)=\int \frac{6 d x}{\sin ^2 x(1-\cot x)^2}=\int \frac{6 \operatorname{cosec}^2 x d x}{(1-\cot x)^2} \) \( \text { Put } 1-\cot x=t \) \( \operatorname{cosec}^2 x d x=d t \) \( I=\int \frac{6 d t}{t^2}=\frac{-6}{t}+c \) \( I(x)=\frac{-6}{1-\cot x} c, c=3 \)…
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