integrate dx/[{(sinx)^3}{sin(x+a)}]^(1/2)

							
Dear student,
Use two things over here
expansion of sin(x+a) and write sin^3x in terms of sin3x...




























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I = dx/[{sin3x}{sinx+a}]1/2dx
sinx+a = sinxcosa + cosxsina
using this integral becomes
I = dx/[{sin3x}{sinxcosa+cosxsina}]1/2
I = dx/[ { sin4xcosa + sin3xcosxsina} ] 1/2
now divide numerator & denominator by cos2x
I = sec2xdx/[ {tan4xcosa + tan3xsina} ]1/2
now put tanx = t , sec2xdx = dt
I = dt/[{t4cosa+t3sina}]1/2
now divide numerator & denominator by t2
I = (1/t2)dt/[ {cosa + sina/t} ]1/2
now put cosa+sina/t = U
         sina(1/t2)dt = -dU
now I becomes
I = (-coseca)dU/ { U }1/2
I = (2coseca) (U)1/2 + C
or
I = (2coseca) {sin(x+a)/sinx}       +    C
 
approve this if u like it