Question

$\int {\sin }^{5}xdx=$

• A. $\frac{-{\sin }^{4}x\cos x}{5}-\frac{4}{15}{\sin }^{2}x\cos x+\frac{8}{15}\cos x+c$
• B. $\frac{-{\sin }^{4}x\cos x}{5}-\frac{4}{15}{\sin }^{2}x\cos x-\frac{8}{15}\cos x+c$
• C. $\frac{-{\sin }^{4}x\cos x}{5}+\frac{4}{15}{\sin }^{2}x\cos x+\frac{8}{15}\cos x+c$
• D. $none$

Answer 1

The correct option is B $\frac{-{\sin }^{4}x\cos x}{5}-\frac{4}{15}{\sin }^{2}x\cos x-\frac{8}{15}\cos x+c$

${\int }^{}{\sin }^{5}xdx$

$={\int }^{}{\sin }^{4}x\sin xdx$

$={\int }^{}{\left({\sin }^{2}x\right)}^2\sin xdx$

$={\int }^{}{\left(1-{\cos }^{2}x\right)}^2\sin xdx$

putting $\cos x=t$

$\Rightarrow \sin xdx=-dt$

$=-{\int }^{}{\left(1-t^2\right)}^{2}dt$

$=-{\int }^{}\left(1+t^4-2t^2\right)dt$

$=-\left[t+\frac{t^5}{5}-\frac{2t^3}{3}\right]+c$

$=-\left[\cos x+\frac{{\cos }^{5}x}{5}-\frac{2}{3}{\cos }^{3}x\right]+c$

$=\frac{2}{3}{\cos }^{3}x-\frac{{\cos }^{5}x}{5}-\cos x+c$

$=\frac{1}{15}\left[10{\cos }^{3}x-3{\cos }^{5}x-15\cos x\right]+c$

$=\frac{1}{15}\left[{\cos }^{3}x\left(10-3{\cos }^{2}x\right)-15\cos x\right]+c$

$=\frac{1}{15}\left[{\cos }^{2}x\left(10-3{\cos }^{2}x\right)\cos x-15\cos x\right]+c$

$=\frac{1}{15}\left[\left(1-{\sin }^{2}x\right)\left(10-3{\cos }^{2}x\right)-15\right]\cos x+c$

$=\frac{1}{15}\left[\left(1-{\sin }^{2}x\right)\left(10-3+3{\sin }^{2}x\right)-15\right]\cos x+c$

$=\frac{1}{15}\left[\left(1-{\sin }^{2}x\right)\left(7+3{\sin }^{2}x\right)-15\right]\cos x+c$

$=\frac{1}{15}\left[7-4{\sin }^{2}x-3{\sin }^{4}x-15\right]\cos x+c$

$=\frac{1}{15}\left[-8-4{\sin }^{2}x-3{\sin }^{4}x\right]\cos x+c$

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