Let I(x)= 6 ^2x(1- x)^2dx.

Question

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

Accepted Answer

$\begin{gathered} I(x)=\int \frac{6}{{\sin }^{2}x(1-\cot x{)}^{2}}dx \ I(x)=\int \frac{6}{(\sin x-\cos x{)}^{2}}dx \ =\int \frac{6{\sec }^{2}x}{(	an x-1{)}^{2}}dx \end{gathered}$

$\begin{gathered} 	an x=t\Rightarrow {\sec }^{2}xdx=dt \ =\int \frac{6dt}{(t-1{)}^{2}} \ =-\frac{6}{(t-1)}+c \ =\frac{-6}{(	an x-1)}+c \ I(x)=\frac{6}{1-	an x}+c \ I(0)=3 \ \frac{6}{1-	an 0}+c=3 \ c=-3 \end{gathered}$

$\begin{gathered} I(x) =\frac{6}{1-	an x}-3 \ I(\frac{\pi }{12}) =\frac{6}{1-	an (\frac{\pi }{12})}-3 \ =\frac{6}{1-(2-\sqrt{3})}-3 \ =\frac{6}{\sqrt{3}-1}-3 \ =\frac{6-3\sqrt{3}+3}{\sqrt{3}-1} \ =\frac{9-3\sqrt{3}}{\sqrt{3}-1} \ =\frac{3\sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1} \ =3\sqrt{3} \end{gathered}$

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