Let I(x) = ∫ 6/sin^2 x(1 - cot)^2 dx. If I(0) = 3, then I(π/ 12) is equal to :

Question

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

(1) \(\sqrt{3}\)

(2) \(3 \sqrt{3}\)

(3) \(6 \sqrt{3}\)

(4) \(2 \sqrt{3}\)

Answer 1

Correct option is (2) \(3 \sqrt{3}\)

\(I(x)=\int \frac{6 d x}{\sin ^{2} x(1-\cot x)^{2}}=\int \frac{6 \operatorname{cosec}^{2} x d x}{(1-\cot x)^{2}}\)

Put \(1-\cot x=t\)

\(\operatorname{cosec}^{2} x d x=d t\)

\(\mathrm{I}=\int \frac{6 \mathrm{dt}}{\mathrm{t}^{2}}=\frac{-6}{\mathrm{t}}+\mathrm{c}\)

\(\mathrm{I}(\mathrm{x})=\frac{-6}{1-\cot \mathrm{x}} \mathrm{c}, \mathrm{c}=3\)

\(I(x)=3-\frac{6}{1-\cot x}, I\left(\frac{\pi}{12}\right)=3-\frac{6}{1-(2+\sqrt{3})}\)

\(I\left(\frac{\pi}{12}\right)=3+\frac{6}{\sqrt{3}+1}=3+\frac{6(\sqrt{3}-1)}{2}=3 \sqrt{3}\)