Question

If $\int {\sin }^{n}dx=-\frac{\cos x{\sin }^{n-1}x}{n}+f\left(n\right)\int si{n}^{n-2}xdx$, then f(3) is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

Let $I_n=\int {\sin }^{n}xdx=\int \sin x{\sin }^{n-1}xdx$
$\therefore$ $I_n={\sin }^{n-1}x(-\cos x)-\int (n-1){\sin }^{n-2}x\left(\cos x\right)(-\cos x)dx$
$=-{\sin }^{n-1}x\cos x+(n-1)\int {\sin }^{n-2}x(1-{\sin }^{2}x)dx$
$=-{\sin }^{n-1}x\cos x+(n-1)\int {\sin }^{n-2}xdx-(n-1)I_n$
$\therefore$ $I_n(1+n-1)=-{\sin }^{n-1}x\cos x+(n-1)\int {\sin }^{n-2}xdx$
$\therefore$ $I_n=-\frac{{\sin }^{n-1}x\cos x}{n}+\left(\frac{n-1}{n}\right)\int {\sin }^{n-2}xdx$
$I_n=\frac{-{\sin }^{n-1}x\cos x}{n}+f\left(n\right)\int {\sin }^{n-2}xdx$
where $f\left(n\right)=\frac{n-1}{n}$
$\therefore$ $f\left(3\right)=\frac{2}{3}$