Question

If $\int si{n}^{5}xdx=-\frac{1}{5}si{n}^{4}xcosx+Asi{n}^{2}xcosx-\frac{8}{15}cosx+C$ then A is equal to

• A. $-2/15$
• B. $-3/5$
• C. $-4/15$
• D. $-1/15$

Answer 1

Answer: (C)

$-4/15$

Let $I=\int {\sin }^{5}xdx$
Using $\int {\sin }^{m}xdx=\frac{-\cos x{\sin }^{m-1}x}{m}+\frac{m-1}{m}\int {\sin }^{-2+m}xdx$
$I=-\frac{1}{5}{\sin }^{4}x\cos x+\frac{4}{5}\int {\sin }^{3}xdx$
$=-\frac{1}{5}{\sin }^{4}x\cos x-\frac{4}{15}{\sin }^{2}x\cos x+\frac{8}{15}\int \sin xdx$
$=-\frac{1}{5}{\sin }^{4}x\cos x-\frac{4}{15}{\sin }^{2}x\cos x+\frac{8}{15}\cos x$