Question

Evaluate :   `∫1/(cos^4x+sin^4x)dx`

Answer 1

Let I = `int 1/(cos^4x+sin^4x) dx`
Divide numerator and denominator by cos⁴x, we get:

`int [sec^4x]/[1 + tan^4x]` dx

`int  [sec^2 x(sec^2x)]/[ 1 + tan^4x ]` dx

`int  [sec^2 x( 1 + tan^2 x)]/( 1 + tan^4x )`dx

Putting tan x = t,
Sec²x dx = dt

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

Dividing the numerator and denominator by t², we get:

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

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

Let t - `1/t` = u

`1 + 1/t^2 = (du)/dt`

`( 1 + 1/t^2) dt = du`

I = `1/sqrt2 tan^-1 (u/sqrt2) + C`

I = `1/sqrt2 tan^-1 (( t - 1/t )/sqrt2) + C`

I = `1/sqrt2 tan^-1 ((t^2 - 1)/(sqrt2t)) + C`

I = `1/sqrt2 tan^-1  ( tan^2x - 1)/(sqrt2tan x ) + C`.