The integral `underset(0)overset(pi)int x f(sinx )dx` is equal to

Options

A. `pioverset(pi)underset(0)int f(cos x)dx`
B. `pioverset(pi)underset(0)int f(sin x)dx`
C. `(pi)/(2)overset(pi//2)underset(0)int f(sin x)dx`
D. `pi overset(pi//2)underset(0)int f(cos x)dx`

Answer

D

Explanation

Let `I=overset(pi)underset(0)int x f(sin x)dx " "`…….(i)
Then,
`I=overset(pi)underset(0)int (pi-x)(sin x)dx`
`rArr I=overset(pi)underset(0)int (pi-x)f(sin x)dx " "`............(ii)
Adding (i) and (ii), we get
`2I=pioverset(pi)underset(0)int f(sin x)dx`
`rArr 2I=2pioverset(pi//2)underset(0)int f(sin x)dx{:[("Using":overset(2a)underset(0)int dx=2overset(a)underset(0)int f(x)dx","),("when "f(2a-x)=f(x)):}]`
`rArr I=pi overset(pi//2)underset(0)int f(sin x)dx`
`rArr I=pioverset(pirr2)underset(0)int f(sin(pi//2-x))dx=pioverset(pi//2)underset(0)int f(cos x)dx`