Question

For $n\ge 2$, if $I_n=\int (\sin x+\cos x{)}^{n}dx$ then $n{I}_n-2(n-1)I_{n-2}=$

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

Answer 1

Answer: (C)

We have,
$I_n=\int (\sin x+\cos x{)}^{n}dx$
$=\int (\sin x+\cos x{)}^{n-1}\cdot (\sin x+\cos x)dx$
$=\int (\sin x+\cos x{)}^{n-1}(\sin x-\cos x)$
$-\int (n-1)(\sin x+\cos x{)}^{n-2}(\cos x-\sin x)(\sin x-\cos x)dx$
Integration by parts,
$\therefore I_n=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)$
$+(n-1)\int (\sin x+\cos x{)}^{n-2}\left\{(\sin x-\cos x{)}^2\right\}dx$
As we know that,
$(\sin x+\cos x{)}^2+(\sin x-\cos x{)}^2=2$
$I_n=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)+(n-1)$
$\int (\sin x+\cos x{)}^{n-2}2-(\sin x+\cos x{)}^{2}dx$
$=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)+(n-1)$
$\int 2(\sin x+\cos x{)}^{n-2}dx-(n-1)\int (\sin x+\cos x)dx$
$=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)+2(n-1)$
$I_{n-2}=(n-1)I_n$
$\Rightarrow n{I}_n-2(n-1)I_{n-2}=(\sin x+\cos x{)}^{n-1}$
$(\sin x-\cos x)+c$