Question

If $I_{m,n}=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{m}x\sin nxdx$, then $7I_{4,3}-4I_{3,2}$ is equal to

• A. $1$
• B. $-1$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is A $1$

$I_{4,3}=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}x\sin 3xdx$
Applying Integrating by parts, we have

$I_{4,3}={[-\frac{\cos 3x{\cos }^{4}x}{3}]}_0^{\pi /2}-\frac{4}{3}\underset{0}{\overset{\pi /2}{\int }}{\cos }^{3}x\sin x\cos 3xdx$
But $\sin x\cos 3x=-\sin 2x+\sin 3x\cos x$
So,
$I_{4,3}=\frac{1}{3}+\frac{4}{3}\underset{0}{\overset{\pi /2}{\int }}{\cos }^{3}x\sin 2xdx-\frac{4}{3}\underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}x\sin 3xdx=\frac{1}{3}+\frac{4}{3}{I}_{3,2}-\frac{4}{3}{I}_{4,3}\Rightarrow \frac{7}{3}{I}_{4,3}-\frac{4}{3}{I}_{3,2}=\frac{1}{3}\therefore 7I_{4,3}-4I_{3,2}=1$