Question

If $m\ne n$ and $I=\int \sin mx\sin nxdx$, then $I$ equals:

• A. $\frac{\sin (m-n)x}{m-n}+\frac{\sin (m+n)x}{m+n}+C$
• B. $\frac{\cos (m-n)x}{m-n}-\frac{\cos (m+n)x}{m+n}+C$
• C. $\frac{1}{2}[\frac{\sin (m-n)x}{m-n}-\frac{\sin (m+n)x}{m+n}]+C$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}[\frac{\sin (m-n)x}{m-n}-\frac{\sin (m+n)x}{m+n}]+C$

Let $I=\int \sin mx\sin nxdx=\frac{1}{2}\int (\cos (m-n)x-\cos (m+n)x)dx$

$=\frac{-1}{2}\int \cos (mx+nx)dx+\frac{1}{2}\int \cos (mx-nx)dx$

Substitute $u=x(m-n)\Rightarrow du=(m-n)dx$

$t=x(m+n)\Rightarrow dt=(m+n)dx$

$\therefore I=-\frac{1}{2(m+n)}\int \cos tdt+\frac{1}{2(m-n)}\int \cos udu$

$=\frac{\sin u}{2(m-n)}-\frac{\sin t}{2(m+n)}+c$

$=\frac{1}{2}[\frac{\sin (m-n)x}{m-n}-\frac{\sin (m+n)x}{m+n}]+c$