Thank you for reporting, we will resolve it shortly
Q.
If $I_{m, n}=\int \cos ^{m} x \sin n x d x$, then $7 I_{4,3}-4 I_{3,2}$ is equal to
Integrals
Solution:
$I_{4,3}=\int \cos ^{4} x \sin 3 x d x$
Integrating by parts, we have
$I_{4,3}=-\frac{\cos 3 x \cos ^{4} x}{3}-\frac{4}{3} \int \cos ^{3} x \sin x \cos 3 x d x$
But $\sin x \cos 3 x=-\sin 2 x+\sin 3 x \cos x$. So,
$I_{4,3}=-\frac{\cos x \cos ^{4} x}{3}+\frac{4}{3} \int \cos ^{3} x \sin 2 x d x$
$-\frac{4}{3} \int \cos ^{4} x \sin 3 x d x+C$
$=-\frac{\cos 3 x \cos ^{4} x}{3}+\frac{4}{3} I_{3,2}-\frac{4}{3} I_{4,3}+C$
Therefore, $\frac{7}{3} I_{4,3}-\frac{4}{3} I_{3,2}=-\frac{\cos 3 x \cos ^{3} x}{3}+C$
or $7 I_{4,3}-4 I_{3,2}=-\cos 3 x \cos ^{4} x+ C$