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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... Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
Dear student,
Use two things over here
expansion of sin(x+a) and write sin^3x in terms of sin3x...
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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
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
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