Consider `I = int_(0)^(pi) (xdx)/(1+sinx)`
What is `int_(0)^(pi) (dx)/(1+sinx)` equal to ?

Options

A. `1`
B. `2`
C. `4`
D. `-2`

Answer

B

Explanation

Given , `I = int_(0)^(pi) (xdx)/(1+sinx) "……."(i)`
`= int_(0)^(pi) ((pi-x))/(1+sin(pi-x)) dx`
`[ :' int_(0)^() f(x) dx = int_(0)^(a) f(a-x) dx]`
`= int_(0)^(pi) ((pi-x))/(1+sinx) dx"…….."(ii)`
`[ :' sin(pi-x) = sin x]`
Adding eqs. (i) and (ii) , we get
`2I = piint_(0)^(pi) (dx)/(1+sinx) "........."(iii)`
`rArr 2I = 2piint_(0)^(pi/2) (dx)/(1+sinx)`
`[ :' int_(0)^(2a) f(x) dx = 2int_(0)^(a) f(x)dx " if " f (2a-x) = f(x)]`
`rArr I = piint_(0)^(pi/2) (dx)/(1+((2tan'x2)/(1+tan^(2)'x/2)))`
`rArr I = piint_(0)^(pi/2) (sec^(2)' x/2dx)/(tan^(2)'x/2+1+2tan 'x/2)`
`= I = piint_(0)^(pi/2) ((sec^(2)'x/2))/((tan'x/2+1)^(2))`
Let `tan 'x/2 + 1 = t`
`rArr sec^(2)'x/2.(1)/2 dx = dt`
`rArr sec^(2)'x/2 dx = 2dt`
When `x = 0`, then `t = 1` and when `x = pi/2`, then `t = 2`
`:. I = 2piint_(1)^(2) (dt)/(t^(2)) = 2pi[(tan^(-2+1))/(-2+1)]_(1)^(2) = - 2pi[1/t]_(1)^(2)`
`= -2pi [1/2 - 1]`
` = - 2pi (-1/2) = pi`
From eq. (iii) .
`2I = piint_(0)^(pi) (dx)/(1+sinx)`
`rArr int_(0)^(pi) (dx)/(1+sinx) = 2/pi I`
`rArr int_(0)^(pi) (dx)/(1+sinx) = 2/pi x pi = 2 " " ( :. I = pi)`