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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?

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  • $\begingroup$ See also here. $\endgroup$ – mrtaurho Jan 14 at 10:45
  • $\begingroup$ 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? $\endgroup$ – Zacky Jan 14 at 11:12
  • $\begingroup$ @Zacky. Decent is a very good word for this problem. Cheers :-) $\endgroup$ – Claude Leibovici Jan 14 at 11:27
  • $\begingroup$ @Zacky I'm not sure of that. WolframAlpha outputs the result here $\endgroup$ – mengdie1982 Jan 14 at 11:30
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    $\begingroup$ Please include some context or background - where did the integral arise, and why is it of interest? $\endgroup$ – Carl Mummert Jan 14 at 12:07

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