# How do you slove int [sin4x/(sin^(8)x+cos^(8)x)] dx

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### 2 Answers

You need to start by expand the square of the basic formula of trigonometry such that:

`(sin^2 x + cos^2 x)^2 = sin^4 x + cos^4 x + 2sin^2 x*cos^2 x`

`1 = sin^4 x + cos^4 x + 2sin^2 x*cos^2 x`

`1 - 2sin^2 x*cos^2 x = sin^4 x + cos^4 x`

`(1 - 2sin^2 x*cos^2 x)^2 =sin^8 x + cos^8 x + 2sin^4 x*cos^4 x`

`1 - 4sin^2 x*cos^2 x + 4sin^4 x*cos^4 x = sin^8 x + cos^8 x + 2sin^4 x*cos^4 x`

You need to substitute `2sin x*cos x = sin 2x`

`1 - sin^2 2x = sin^8 x + cos^8 x - 2sin^4 x*cos^4 x`

`cos^2 2x = sin^8 x + cos^8 x - 2sin^4 x*cos^4 x`

`sin^8 x + cos^8 x = cos^2 2x + 2sin^4 x*cos^4 x`

You need to write the numerator considering the sine of double angle formula such that:

`sin 4x = 2sin 2xcos2x`

You should come up with the substitution such that:

`cos 2x = t => -2sin 2x dx = dt`

`sin x*cos x = (sin 2x)/2`

`2sin^4 x*cos^4 x = (sin^4(2x))/8 = (1 - cos^2(2x))^2/8`

`2sin^4 x*cos^4 x = (1-t^2)^2/8`

`int (sin 4x)/(cos^2 2x + 2sin^4 x*cos^4 x) dx= int -tdt/(t^2 + (1-t^2)^2/8)`

`int -8tdt/(8t^2 + 1 - 2t^2 + t^4) = -8 int tdt/(t^4 + 6t^2 + 1)`

`-8 int tdt/(t^4 + 6t^2 + 1) = -8 int tdt/(t^4 + 6t^2 +9 - 9+ 1)`

`-8 int tdt/(t^4 + 6t^2 + 1) = -8 int tdt/((t^2+3)^2 - 8) `

You need to substitute u for `t^2+3` such that:

t^2+3 = u => 2tdt = du

`-4 int 2tdt/((t^2+3)^2 - 8) = -4 int (du)/(u^2 - 8)`

`-4 int (du)/(u^2 - 8) = -2/(sqrt8)*ln|(u-sqrt8)/(u+sqrt8)| + c`

You need to substitute `t^2+3` for u such that:

`-4 int 2tdt/((t^2+3)^2 - 8) =-2/(sqrt8)*ln|((t^2+3)-sqrt8)/((t^2+3)+sqrt8)| + c`

You need to substitute `cos 2x` for t such that:

`int (sin 4x)/(cos^2 2x + 2sin^4 x*cos^4 x) dx = -2/(sqrt8)*ln|((cos^2 2x+3)-sqrt8)/((cos^2 2x+3)+sqrt8)| + c`

**Hence, evaluating the given trigonometric integral yields:**

`int (sin 4x)/(cos^2 2x + 2sin^4 x*cos^4 x) dx = -2/(sqrt8)*ln|((cos^2 2x+3)-sqrt8)/((cos^2 2x+3)+sqrt8)| + c`

Thank you very much.

I did a few more steps and I came to 1/(sqrt)*ln|(cos4x+7+4(sqrt2))/(cos4x+7-4(sqrt2))| +c