Question

If $I_n={\int }_0^{\pi /2}{(\sin x+\cos x)}^{n}dx$, then value of $n{I}_n-2(n-1)I_{n-2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $n$

Answer 1

The correct option is C $2$

Let

$n=2$
Then

$n{I}_n-2(n-1)I_{n-2}$
$=2I_2-2I_0$
Now

$I_0={\int }_0^{\frac{\pi }{2}}\left(cos\right(x)+sin(x){)}^0.dx$

$=\frac{\pi }{2}$

And

$I_2={\int }_0^{\frac{\pi }{2}}\left(sin\right(x)+cos(x){)}^2.dx$

$I_2={\int }_0^{\frac{\pi }{2}}1+sin2x$

$=\frac{\pi }{2}+1$

Hence

$2(I_2-I_0)$
$=2(\frac{\pi }{2}+1-\frac{\pi }{2})$
$=2$

Hence correct option is C.