Question

If $I_n=\int {\sin }^{n}xdx,$, then ${nI}_n-(n-1)I_{n-2}=$

• A. ${\sin }^{n-1}x\cos x$
• B. ${\cos }^{n-1}x\sin x$
• C. $-{\sin }^{n-1}x\cos x$
• D. $-{\cos }^{n-1}x\sin x$

Answer 1

The correct option is C $-{\sin }^{n-1}x\cos x$

$I_n={\int }^{}{\sin }^{n}xdx$

$={\int }^{}{\sin }^{n-1}x\sin xdx$

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

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

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

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

$\Rightarrow I_n=-{\sin }^{n-1}x\cos x+\left(n-1\right)I_{n-2}-\left(n-1\right)I_n$

$\Rightarrow n{I}_n=-{\sin }^{n-1}x\cos x+\left(n-1\right)I_{n-2}$

$\Rightarrow n{I}_n-\left(n-1\right)I_{n-2}=-{\sin }^{n-1}x\cos x$