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`        integral 1/(sin^3x+cos^3x) dx`
7 years ago

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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