Question

 

 

Evaluate :

`int1/(sin^4x+sin^2xcos^2x+cos^4x)dx`

 

 

Answer 1

We need to evaluate `int1/(sin^4x+sin^2xcos^2x+cos^4x)dx`

`Let I=int1/(sin^4x+sin^2xcos^2x+cos^4x)dx`

Multiply the numerator and the denominator by sec⁴x, we have

`I=int(sec^4dx)/(tan^4x+tan^2x+1)`

`I=int(sec^2x xx sec^2x dx)/(tan^4s+tan^2x+1)`

Now substitute t=tanx;dt=sec²xdx

Therefore,

`I=int(1+t^2)/(t^4+t^2+1)dt`

`I=int(1+1/t^2)/(t^2+1/t^2+1)dt`

`I=int(1+1/t^2)/(t^2+1/t^2-2+2+1)dt`

`I=int(1+1/t^2)/((T-1/T)^2+3)dt`

Substitute `z=t-1/t; dz=(1+1/t^2)dt`

`I=int(dz)/(z^2+3)`

`I=int(dz)/(z^2+(sqrt3)^2)`

`I=1/sqrt3 tan^(-1)(z/sqrt3)+c`

`I=1/sqrt3tan^(-1)((t-1/t)/sqrt3)+c`

`I=1/sqrt3tan^(-1)((tanx-1/tanx)/sqrt3)+c`

`I=1/sqrt3tan^(-1)((tanx-cotx)/sqrt3)+c`