I got a different answer to it.
I am getting I = (pi)2 / 2 sin(a).
nyhow approach is like dis.
First apply property int x from 0 to a [ f(x) ] = int x from 0 to a [ f(a - x) ] and then apply
int x from 0 to 2a [ f(x) ] = 2 int x from 0 to a [ f(x) ] , if f(2a - x) = f(x)
U'll get,
I = (pi) int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx
Let I1 = int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx
Now put cos(a) sin(x) = y and u'll get
I1 = int y from 0 to cos(a) ( 1 / (1 - y) (cos2a - y2)1/2 ).dy
Now put cos2a - y2 = 1/t
U will now get, I1 = (1/sina) int t from 1 to (1 / 1 - cos(a)) [ (cos(a)/sin2(a))2 - (t - (1/sin2a))2 ]1/2
So, finally I1 = pi / 2 sin(a)
So, I = (pi)2 / 2 sin(a) [Ans]..............
I got a different answer to it.
I am getting I = (pi)2 / 2 sin(a).
nyhow approach is like dis.
First apply property int x from 0 to a [ f(x) ] = int x from 0 to a [ f(a - x) ] and then apply
int x from 0 to 2a [ f(x) ] = 2 int x from 0 to a [ f(x) ] , if f(2a - x) = f(x)
U'll get,
I = (pi) int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx
Let I1 = int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx
Now put cos(a) sin(x) = y and u'll get
I1 = int y from 0 to cos(a) ( 1 / (1 - y) (cos2a - y2)1/2 ).dy
Now put cos2a - y2 = 1/t
U will now get, I1 = (1/sina) int t from 1 to (1 / 1 - cos(a)) [ (cos(a)/sin2(a))2 - (t - (1/sin2a))2 ]1/2
So, finally I1 = pi / 2 sin(a)
So, I = (pi)2 / 2 sin(a) [Ans]..............