Question

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

• A. constant
• B. $-{\cos }^{2}x+C$
• C. $-{\cos }^{4}x\cos 3x+C$
• D. $\cos 7x-\cos 4x+C$

Answer 1

Answer: (C)

$I_{4,3}=\int {\cos }^{4}x\sin 3xdx$
Integrating by parts, we have
$I_{4,3}=-\frac{\cos 3x{\cos }^{4}x}{3}-\frac{4}{3}\int {\cos }^{3}x\sin x\cos 3xdx$
But $\sin x\cos 3x=-\sin 2x+\sin 3x\cos x$. So,
$I_{4,3}=-\frac{\cos x{\cos }^{4}x}{3}+\frac{4}{3}\int {\cos }^{3}x\sin 2xdx$
$-\frac{4}{3}\int {\cos }^{4}x\sin 3xdx+C$
$=-\frac{\cos 3x{\cos }^{4}x}{3}+\frac{4}{3}{I}_{3,2}-\frac{4}{3}{I}_{4,3}+C$
Therefore, $\frac{7}{3}{I}_{4,3}-\frac{4}{3}{I}_{3,2}=-\frac{\cos 3x{\cos }^{3}x}{3}+C$
or $7I_{4,3}-4I_{3,2}=-\cos 3x{\cos }^{4}x+C$