Day 16: The Powers of Trigonometric Integrals

It would be a nightmare to type up this solution so here is a written one:

And therefore the answer is $\boxed{37}$ .

As Michael put in, it would be a nightmare to type up the solution.

Let me present an outline.

Write $\large \cos^4\theta$ + $\large \sin^4\theta$ =( $\large \cos^2\theta$ + $\large \sin^2\theta$ ) ^2 - $\large \ 2cos^2\theta$ $\large \ 2sin^2\theta$ = 1 - $\large \ 2cos^2\theta$ $\large \ 2sin^2\theta$

Open up the brackets to obtain an expression of the form $\large \sin^8\theta$ + $\large \cos^8\theta$ - $\large \ 2cos^4\theta$ $\large \ 2sin^4\theta$ + $\large \ cos^2\theta$ $\large \ sin^2\theta$ = ( $\large \cos^4\theta$ - $\large \sin^4\theta$ )^2 + $\large \ cos^2\theta$ $\large \ sin^2\theta$ .

Now apply standard trigonometric identities to obtain the integral.