∫dx/[sin(x/2)+Cos(x/2)]=?

I = dx/[ sinx/2 + cosx/2 ]

 =dx/2^1/2[(sinx/2)(1/2^1/2) + (cosx/2)(1/2)^1/2]                   (multiplying dividing by 2^1/2)

 =dx/2^1/2[(sinx/2)(cospi/4) + (cosx/2)(sinpi/4)]

 =dx/2^1/2[sin(x/2+pi/4)]                                        (sina+b = sinacosb+cosasinb)

=cosec(x/2+pi/4)dx/2^1/2

=log[cosec(x+pi/4) -cot(x+pi/4)]  + c

Approved