Question

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

• 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

$-{\sin }^{n-1}x\cos x$

Explanation for the correct option.

Step 1: Solve $I_n=\int {\sin }^{n}xdx$.

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

Integrating by part, we get

$\begin{array}{rcl}{I}_n& =& {\sin }^{n-1}x\left(-\cos x\right)-\int \left(n-1\right){\sin }^{n-2}x·\cos x·-\cos xdx\\ & =& -{\sin }^{n-1}x·\cos x+\int \left(n-1\right){\sin }^{n-2}x·{\cos }^{2}xdx\\ & =& -{\sin }^{n-1}x·\cos x+\int \left(n-1\right){\sin }^{n-2}x·\left(1-{\sin }^{2}x\right)dx\\ & =& -{\sin }^{n-1}x·\cos x+\int \left(n-1\right)\left({\sin }^{n-2}x-{\sin }^{n}x\right)dx\\ & =& -{\sin }^{n-1}x·\cos x+\left(n-1\right)\int {\sin }^{n-2}xdx-\left(n-1\right)\int {\sin }^{n}xdx\\ & =& -{\sin }^{n-1}x·\cos x+\left(n-1\right)I_{n-2}-\left(n-1\right)I_n....\left(1\right)\end{array}$

Step 2: Find the value of $n{I}_n-(n-1)I_{n-2}$.

From $\left(1\right)$, we get

$\begin{array}{rcl}{I}_n& =& -{\sin }^{n-1}x·\cos x+\left(n-1\right)I_{n-2}-\left(n-1\right)I_n\\ & \Rightarrow & I_n-\left(n-1\right)I_{n-2}+\left(n-1\right)I_n=-{\sin }^{n-1}x·\cos x\\ \Rightarrow \left(n-1+1\right)I_n-\left(n-1\right)I_{n-2}& =& -{\sin }^{n-1}x·\cos x\\ \Rightarrow n{I}_n-\left(n-1\right)I_{n-2}& =& -{\sin }^{n-1}x·\cos x\end{array}$

Hence, option C is correct.