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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Find the integral: dx/(cosx+sinx)^(1/2)
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mohamed arafath (20)

Olaaa!! Perrrfect answer. 4  [4 rates]

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put t^2=(cosx+sinx)             (cosx-sinx)=2tdt/dx              sin2x=t^4-1         (cosx-sinx)^2=1-sin2x                                    cosx-sinx={2-t^4}^1/2     dx/(cosx+sinx)^1/2 = 2dt/{2-t^4}^1/2 i hope after u can solve this

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