Let $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. If $I(0)=3$, then $I\left(\frac{\pi}{12}\right)$ is equal to

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Aicharya · Accepted Answer

$I(x)=\int \frac{6 d x}{\sin ^2 x(1-\cot x)^2}=\int \frac{6 \operatorname{cosec}^2 x d x}{(1-\cot x)^2}$
Put $1-\cot x=t$ $\operatorname{cosec}^2 \mathrm{xdx}=\mathrm{dt}$ $\begin{aligned} & I=\int \frac{6 d t}{t^2}=\frac{-6}{t}+c \\ & I(x)=\frac{-6}{1-\cot x} c, c=3 \\ & I(x)=3-\frac{6}{1-\cot x}, I\left(\frac{\pi}{12}\right)=3-\frac{6}{1-(2+\sqrt{3})} \\ & I\left(\frac{\pi}{12}\right)=3+\frac{6}{\sqrt{3}+1}=3+\frac{6(\sqrt{3}-1)}{2}=3 \sqrt{3}\end{aligned}$

Let $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. If $I(0)=3$, then $I\left(\frac{\pi}{12}\right)$ is…

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