Question

Which of the following is/are equal to $\int {\sin }^{5}xdx$
(where $C$ is integration constant)

• A. $\frac{1}{5}\cos x{\sin }^{4}x+\frac{4}{15}\cos x-\frac{4}{5}\cos x+C$
• B. $-\frac{{\cos }^{5}x}{5}+\frac{2{\cos }^{3}x}{3}-\cos x+C$
• C. $-\frac{1}{5}\cos x{\sin }^{4}x-\frac{4}{15}\cos x{\sin }^{2}x-\frac{8}{15}\cos x+C$
• D. $\frac{{\cos }^{5}x}{5}+\frac{2{\cos }^{3}x}{3}-\cos x+C$

Answer 1

Answer: (B), (C)

$-\frac{1}{5}\cos x{\sin }^{4}x-\frac{4}{15}\cos x{\sin }^{2}x-\frac{8}{15}\cos x+C$

$I=\int {\sin }^{5}xdx=\int {\sin }^{4}x\sin xdx=\int {(1-{\cos }^{2}x)}^2\sin xdx$
Let, $t=\cos x$
$\Rightarrow dt=-\sin xdx=-\int {(1-t^2)}^{2}dt=-\int (1+t^4-2t^2)dt=-[t+\frac{t^5}{5}-\frac{2t^3}{3}]+c=-\cos x-\frac{{\cos }^{5}x}{5}+\frac{2{\cos }^{3}x}{3}+c=-\frac{{\cos }^{5}x}{5}+\frac{2{\cos }^{3}x}{3}-\cos x+c$
Now,
$I_n=\int {\sin }^{n}xdx=-\frac{\cos x{\sin }^{n-1}x}{n}+\frac{n-1}{n}{I}_{n-2}$
$\therefore I_5=-\frac{\cos x{\sin }^{4}x}{5}+\frac{4}{5}{I}_3=-\frac{\cos x{\sin }^{4}x}{5}+\frac{4}{5}[-\frac{\cos x{\sin }^{2}x}{3}+\frac{2}{3}{I}_1]=-\frac{\cos x{\sin }^{4}x}{5}-\frac{4}{15}\cos x{\sin }^{2}x+\frac{8}{15}\int \sin xdx=-\frac{1}{5}\cos x{\sin }^{4}x-\frac{4}{15}\cos x{\sin }^{2}x-\frac{8}{15}\cos x+c^{\prime }$