Simplify the integrand
We have
I ( x ) = ∫ 6 sin 2 x ( 1 − cot x ) 2 d x . I(x)=\int \frac{6}{\sin^2 x\,(1-\cot x)^2}\,dx. I ( x ) = ∫ sin 2 x ( 1 − cot x ) 2 6 d x . Use
cot x = cos x sin x . \cot x=\frac{\cos x}{\sin x}. cot x = sin x cos x . Then
1 − cot x = 1 − cos x sin x = sin x − cos x sin x . 1-\cot x = 1-\frac{\cos x}{\sin x} = \frac{\sin x-\cos x}{\sin x}. 1 − cot x = 1 − sin x cos x = sin x sin x − cos x . So,
( 1 − cot x ) 2 = ( sin x − cos x ) 2 sin 2 x . (1-\cot x)^2 = \frac{(\sin x-\cos x)^2}{\sin^2 x}. ( 1 − cot x ) 2 = sin 2 x ( sin x − cos x ) 2 . Hence the denominator becomes
sin 2 x ( 1 − cot x ) 2 = sin 2 x ⋅ ( sin x − cos x ) 2 sin 2 x = ( sin x − cos x ) 2 . \sin^2 x\,(1-\cot x)^2 = \sin^2 x\cdot \frac{(\sin x-\cos x)^2}{\sin^2 x} = (\sin x-\cos x)^2. sin 2 x ( 1 − cot x ) 2 = sin 2 x ⋅ sin 2 x ( sin x − cos x ) 2 = ( sin x − cos x ) 2 . Therefore,
I ( x ) = ∫ 6 ( sin x − cos x ) 2 d x . I(x)=\int \frac{6}{(\sin x-\cos x)^2}\,dx. I ( x ) = ∫ ( sin x − cos x ) 2 6 d x .
Rewrite using a standard identity
Recall
sin x − cos x = 2 sin ( x − π 4 ) . \sin x-\cos x = \sqrt2\sin\left(x-\frac{\pi}{4}\right). sin x − cos x = 2 sin ( x − 4 π ) . Thus,
( sin x − cos x ) 2 = 2 sin 2 ( x − π 4 ) . (\sin x-\cos x)^2 = 2\sin^2\left(x-\frac{\pi}{4}\right). ( sin x − cos x ) 2 = 2 sin 2 ( x − 4 π ) . So the integrand becomes
6 ( sin x − cos x ) 2 = 6 2 sin 2 ( x − π 4 ) = 3 csc 2 ( x − π 4 ) . \frac{6}{(\sin x-\cos x)^2} = \frac{6}{2\sin^2\left(x-\frac{\pi}{4}\right)} = 3\csc^2\left(x-\frac{\pi}{4}\right). ( sin x − cos x ) 2 6 = 2 sin 2 ( x − 4 π ) 6 = 3 csc 2 ( x − 4 π ) . Hence,
I ( x ) = ∫ 3 csc 2 ( x − π 4 ) d x . I(x)=\int 3\csc^2\left(x-\frac{\pi}{4}\right)\,dx. I ( x ) = ∫ 3 csc 2 ( x − 4 π ) d x .
Integrate
Using
∫ csc 2 u d u = − cot u + C , \int \csc^2 u\,du = -\cot u + C, ∫ csc 2 u d u = − cot u + C , we get
I ( x ) = − 3 cot ( x − π 4 ) + C . I(x) = -3\cot\left(x-\frac{\pi}{4}\right)+C. I ( x ) = − 3 cot ( x − 4 π ) + C .
Use the condition I ( 0 ) = 3 I(0)=3 I ( 0 ) = 3
Substitute x = 0 x=0 x = 0
I ( 0 ) = − 3 cot ( − π 4 ) + C . I(0)=-3\cot\left(-\frac{\pi}{4}\right)+C. I ( 0 ) = − 3 cot ( − 4 π ) + C . Since
cot ( − π 4 ) = − 1 , \cot\left(-\frac{\pi}{4}\right)=-1, cot ( − 4 π ) = − 1 , we get
I ( 0 ) = − 3 ( − 1 ) + C = 3 + C . I(0)=-3(-1)+C=3+C. I ( 0 ) = − 3 ( − 1 ) + C = 3 + C . Given I ( 0 ) = 3 I(0)=3 I ( 0 ) = 3
3 + C = 3 ⟹ C = 0. 3+C=3 \implies C=0. 3 + C = 3 ⟹ C = 0. So,
I ( x ) = − 3 cot ( x − π 4 ) . I(x)=-3\cot\left(x-\frac{\pi}{4}\right). I ( x ) = − 3 cot ( x − 4 π ) .
Evaluate at x = π 12 x=\frac{\pi}{12} x = 12 π
I ( π 12 ) = − 3 cot ( π 12 − π 4 ) = − 3 cot ( − π 6 ) . I\left(\frac{\pi}{12}\right)=-3\cot\left(\frac{\pi}{12}-\frac{\pi}{4}\right)
=-3\cot\left(-\frac{\pi}{6}\right). I ( 12 π ) = − 3 cot ( 12 π − 4 π ) = − 3 cot ( − 6 π ) . Now,
cot ( − π 6 ) = − cot ( π 6 ) = − 3 . \cot\left(-\frac{\pi}{6}\right)=-\cot\left(\frac{\pi}{6}\right)=-\sqrt3. cot ( − 6 π ) = − cot ( 6 π ) = − 3 . Therefore,
I ( π 12 ) = − 3 ( − 3 ) = 3 3 . I\left(\frac{\pi}{12}\right)=-3(-\sqrt3)=3\sqrt3. I ( 12 π ) = − 3 ( − 3 ) = 3 3 .
Check options
The value is
3 3 , 3\sqrt3, 3 3 , which corresponds to Option D .