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

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

Use two things over here

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

All the best.

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Sagar Singh

B.Tech, IIT Delhi

509 Points
13 years ago

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