Q.Integrate with respect to $\theta$: $\int \frac{\sin^6\theta+\cos^6\theta}{\sin^2\theta\cos^2\theta}\,d\theta$

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=\sin^2\theta,\,b=\cos^2\theta$ (so $a+b=1$): $\sin^6\theta+\cos^6\theta=\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta$. Also $\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2\sin^2\theta\cos^2\theta$, so $\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$. Dividing by $\sin^2\theta\cos^2\theta$: the integrand is $\frac{1}{\sin^2\theta\cos^2\theta …

Sum-of-cubes factoring plus Pythagorean identity to simplify the numerator, then spl …

Forgetting the identity $\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta$ and instead leaving the numerator un …