Evaluate : `int_0^pix/(1+sinalphasinx)dx`

Explanation

Given,I=`int_0^pix/(1+sinalphasinx)dx`

I=`int_0^pi(pi-x)/(1+sinalphasin(pi-x))dx`
I=`int_0^pi(pi)/(1+sinalphasinx)dx-int_0^pix/(1+sinalphasinx)dx`

I=`int_0^pi(pi)/(1+sinalphasinx)dx`-I

2I=`int_0^pi(pi)/(1+sinalphasinx)dx`

Put`sinx=(2tan(x/2))/(1+tan^2(x/2))`

2I=`pi int_0^pi((1+tan^2(x/2)))/((1+tan^2(x/2))+sinalpha(2tan(x/2)))dx`

I=`pi/2 int_0^pi((sec^2(x/2)))/((1+tan^2(x/2))+sinalpha(2tan(x/2)))dx`

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