Question

If $I_n=\int {(\sin x+\cos x)}^{n}dx$ and $I_n=\frac{1}{n}{(\sin x+\cos x)}^{n-1}(\sin x-\cos x)+\frac{2k}{n}{I}_{n-2}$ then $k=$

• A. $(n+1)$
• B. $(n-1)$
• C. $(2n+1)$
• D. $(2n-1)$

Answer 1

The correct option is B $(n-1)$

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

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

Integrating by parts

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

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

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

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

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

$⟹k=n-1$