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Use two things over here

expansion of sin(x+a) and write sin^3x in terms of sin3x...

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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