# Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sin^8 t+\cos^8 t}{\sin t+\cos t+7}{\rm d}t$

Problem

Evaluate $$\int_0^{\frac{\pi}{2}}\frac{\sin^8 t+\cos^8 t}{\sin t+\cos t+7}{\rm d}t.$$

Attempt

Let $$x:=\dfrac{\pi}{2}-t$$. Then $$t=\dfrac{\pi}{2}-x$$ and $${\rm d}t=-{\rm d}x.$$ Thus \begin{align*} \int_0^{\frac{\pi}{2}}\frac{\sin^8 t}{\sin t+\cos t+7}{\rm d}t&=-\int_{\frac{\pi}{2}}^0\dfrac{\sin^8\left(\dfrac{\pi}{2}-x\right)}{\sin\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}-x\right)+7}{\rm d}x\\ &=\int_0^{\frac{\pi}{2}}\dfrac{\cos^8 x}{\cos x+\sin x+7}{\rm d}x.\\ \end{align*} Therefore $$\int_0^{\frac{\pi}{2}}\frac{\sin^8 t+\cos^8 t}{\sin t+\cos t+7}{\rm d}t=2\int_0^{\frac{\pi}{2}}\frac{\sin^8 t}{\sin t+\cos t+7}{\rm d}t=2\int_0^{\frac{\pi}{2}}\frac{\cos^8 t}{\sin t+\cos t+7}{\rm d}t.$$ Can we go on from here?

• With $\tan t =x$ we get: $$2\int_0^\infty \frac{x^8}{(1+x^2)^5}\frac{1}{7+\frac{x+1}{\sqrt{x^2+1}}}dx$$ Is there a reason to expect this to have a decent closed form? – Zacky Jan 14 at 11:12