Let [I(x)= (6)/( ^2 x(1- x)^2) d x].

Question

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

Accepted Answer

$$\begin{aligned}
& I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x \
& I(x)=\int \frac{6}{(\sin x-\cos x)^2} d x \
& =\int \frac{6 \sec ^2 x}{(\tan x-1)^2} d x
\end{aligned}$$

$$\begin{aligned}
& \text { Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \
& =\int \frac{6 d t}{(t-1)^2} \
& =-\frac{6}{(t-1)}+c \
& =\frac{-6}{(\tan x-1)}+c \
& I(x)=\frac{6}{1-\tan x}+c \
& I(0)=3 \
& \frac{6}{1-\tan 0}+c=3 \
& c=-3
\end{aligned}$$

$$\begin{aligned}
I(x) & =\frac{6}{1-\tan x}-3 \
I\left(\frac{\pi}{12}\right) & =\frac{6}{1-\tan \left(\frac{\pi}{12}\right)}-3 \
& =\frac{6}{1-(2-\sqrt{3})}-3 \
& =\frac{6}{\sqrt{3}-1}-3 \
& =\frac{6-3 \sqrt{3}+3}{\sqrt{3}-1} \
& =\frac{9-3 \sqrt{3}}{\sqrt{3}-1} \
& =\frac{3 \sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1} \
& =3 \sqrt{3}
\end{aligned}$$

Solutions · 1