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

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

Grade:11

1 Answers

siddhu bala
16 Points
11 years ago

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


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