Let  $I_n=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\left(\ln \left|\sin x\right|\right)\cos \left(2nx\right)dx$ where  $n\in N$. If  $(12I_3-16I_2)$ is  $k\pi$, then value of k is

$I_n=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\underset{I}{\ln \left(\left|\sin x\right|\right)}\underset{II}{\cos \left(2nx\right)}dx$

applying integration by parts

$$\begin{gathered} I_n=\ln \left|\sin x\right|.{\frac{\sin 2nx}{2n}|}_0^{\frac{3\pi }{2}}-\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\frac{\cot x.\sin 2nx}{2n}dx \\ {I}_n=0-\frac{1}{2n}{I}_n^{\text{'}} \\ {I}_n^{\text{'}}=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\frac{\cos x.\sin 2nx}{\sin x}dx \\ {I}_n^{\text{'}}-I_{n-1}^{\text{'}}=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\frac{\cos x(\sin 2nx-\sin (2n-2))x}{\sin x}dx =\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\frac{2\cos x.\cos (2n-1)x\sin x}{\sin x}dx \\ {I}_n^{\text{'}}-I_{n-1}^{\text{'}}=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}2\cos (2n-1)x.\cos xdx=0 \\ {I}_n^{\text{'}}=I_{n-1}^{\text{'}}=I_{n-2}^{\text{'}}=\mathrm{...........}{I}_1^{\text{'}} \end{gathered}$$

$$\begin{gathered} I_n^{\text{'}}=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}\frac{\sin 2x\cos x}{\sin x}dx=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}2{\cos }^{2}xdx=\underset{0}{\overset{\frac{3\pi }{2}}{\int }}(1+\cos 2x)dx=\frac{3\pi }{2} \\ {I}_n=\frac{-3\pi }{4n} \\ 12I_3=-3\pi \\ 16I_2=-6\pi \Rightarrow 12I_3-16I_2=3\pi \end{gathered}$$