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tushar a Grade: 11
        

integral 1/(sin^3x+cos^3x) dx

7 years ago

Answers : (1)

siddhu bala
16 Points
										

sin^3x+cos^3x=(sinx+cosx)(sin^2x+cos^2x-sinxcosx)=(sinx+cosx)(1-(sinx+cosx)^2)/2


since,


2sinxcosx=(sin2x)=(sinx+cosx)^2-1  =   1-(sinx-cosx)^2.


hence,


putting (sinx+cosx)= z and performing partial fractions


we get,  2[(1/z) + (z/(1-z^2))]


 


 


hence integral I=2[ int{1/(sinx+cosx) + (sinx+cosx)/(1-(sinx+cosx)^2)}]


now rewrite (sinx+cosx)^2 =2-(sinx-cosx)^2 in I2


and (sinx+cosx)=rt(2)[sin(45+x)] in I1


 


we get,


I = -rt(2)ln[cosec(45+x)+cot(45+x)]-ln[{1-(sinx-cosx)}/{1+(sinx-cosx)}] +c



7 years ago
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