Question

If $I_n=\int \frac{{\sin }^{2}nx}{{\sin }^{2}x}dx,$ then the value of $I_{n+1}-2I_n+I_{n-1}$ is:
(where $n\in N$ and $C$ is integration constant)

• A. $\frac{1}{n}\sin 2nx+C$
• B. $-\frac{1}{n}\sin 2nx+C$
• C. $\frac{1}{n}\cos 2nx+C$
• D. $-\frac{1}{n}\cos 2nx+C$

Answer 1

The correct option is A $\frac{1}{n}\sin 2nx+C$

Given: $I_n=\int \frac{{\sin }^{2}nx}{{\sin }^{2}x}dx$
$\Rightarrow I_{n+1}-2I_n+I_{n-1}=\int \frac{{\sin }^2(n+1)x-2{\sin }^{2}nx+{\sin }^2(n-1)x}{{\sin }^{2}x}dx$
$=\int \frac{\left({\sin }^2\right(n+1)x-{\sin }^{2}nx)-({\sin }^{2}nx-{\sin }^2(n-1\left)x\right)}{{\sin }^{2}x}dx$
$=\int \frac{\left(\sin \right(2n+1)x\cdot \sin x)-\left(\sin \right(2n-1)x\cdot \sin x)}{{\sin }^{2}x}dx(\because {\sin }^{2}A-{\sin }^{2}B=\sin (A+B\left)\sin \right(A-B\left)\right)=\int \frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx$
$=\int \frac{2\sin x\cdot \cos 2nx}{\sin x}dx$
$=2\int \cos 2nxdx=\frac{1}{n}\sin 2nx+C$