Question

If $\int {\sin }^{5}xdx=-\frac{1}{5}{\sin }^{4}x\cos x+A{\sin }^{2}x\cos x-\frac{8}{15}\cos x+c$ then A is equal to

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

Answer 1

Answer: (C)

-4/15

Given, $\int {\sin }^{5}xdx=-\frac{1}{5}{\sin }^{4}x\cos x+A{\sin }^{2}x\cos x-\frac{8}{15}\cos x+c$ ....(1)
We have reduction formula for
$\int {\sin }^{n}xdx=-\frac{\cos x{\sin }^{n-1}x}{n}+\frac{n-1}{n}\int {\sin }^{n-2}xdx$
So, $\int {\sin }^{5}xdx=-\frac{\cos x{\sin }^{4}x}{5}+\frac{4}{5}\int {\sin }^{3}xdx$
$=-\frac{\cos x{\sin }^{4}x}{5}+\frac{4}{5}\int (\frac{3\sin x}{4}-\frac{\sin 3x}{4})dx$
$=-\frac{\cos x{\sin }^{4}x}{5}+\frac{3}{5}\int \sin xdx-\frac{1}{5}\int \sin 3xdx$
$=-\frac{\cos x{\sin }^{4}x}{5}-\frac{3}{5}\cos x+\frac{1}{15}\cos 3x+C$
$\int {\sin }^{5}xdx=-\frac{\cos x{\sin }^{4}x}{5}-\frac{3}{5}\cos x+\frac{1}{15}(4{\cos }^{3}x-3\cos x)+C$
$\int {\sin }^{5}xdx=-\frac{\cos x{\sin }^{4}x}{5}-\frac{4}{5}\cos x+\frac{4}{15}(1-{\sin }^{2}x)\cos x+C$
$\int {\sin }^{5}xdx=-\frac{\cos x{\sin }^{4}x}{5}-\frac{8}{15}\cos x-\frac{4}{15}{\sin }^{2}x\cos x+C$
On comparing with (1)
$\Rightarrow A=-\frac{4}{15}$