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Kuldeep Mukherjee Grade: 12
        integrate dx/[{(sinx)^3}{sin(x+a)}]^(1/2)
6 years ago

Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


Use two things over here


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






























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Sagar Singh


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6 years ago
vikas askiitian expert
510 Points
										

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

6 years ago
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