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I'm trying to find the mass of a spherical object with a given density function, and to do so I must solve this integral $$\int_{0}^{\pi}\sin^5{\theta}\cos^2{\theta}\ d\theta,$$

but no matter which method I choose (integration by parts, substitution, etc) I can't for the life of my figure out the anti-derivative.


marked as duplicate by Hans Lundmark, Michael Albanese, drhab, Davide Giraudo, Rick Decker Jun 13 '14 at 15:35

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  • 2
    $\begingroup$ $\sin^5\theta\cos^2\theta=\sin\theta (1-\cos^2\theta)^2\cos^2\theta$. Let $u=\cos\theta$. $\endgroup$ – David Mitra Jun 13 '14 at 14:03
  • $\begingroup$ Who cares about the anti-derivative? The problem doesn't ask for one. $\endgroup$ – David H Jun 13 '14 at 14:09
  • $\begingroup$ See also Wallis' integrals. $\endgroup$ – Lucian Jun 13 '14 at 18:02

Let $x=\cos\theta$ then $dx=-\sin\theta d\theta$ hence

$$\int_0^\pi \sin^5\theta\cos^2\theta d\theta=\int_{-1}^1(1-x^2)^2x^2dx$$ Can you take it from here?

  • $\begingroup$ Yes I can thank you. I didn't think of using trigonometric identities, silly me. $\endgroup$ – Oria Gruber Jun 13 '14 at 14:05
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jun 13 '14 at 14:06
  • $\begingroup$ Are you sure there shouldnt be a $-$ sign there? Since $x=\cos\theta$, then $dx=-\sin\theta d\theta$ $\endgroup$ – Oria Gruber Jun 13 '14 at 14:11
  • 1
    $\begingroup$ By this sign we interchange the limits of the integral: $$-\int_1^{-1}=\int_{-1}^1$$ $\endgroup$ – user63181 Jun 13 '14 at 14:13
  • $\begingroup$ Ah yes that is true. $\endgroup$ – Oria Gruber Jun 13 '14 at 14:17

Another approach:

Consider Beta function $$ \text{B}(x,y)=2\int_0^{\Large\frac\pi2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\ d\theta=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}. $$ Rewrite $$ \int_0^{\large\pi}\sin^5\theta\cos^2\theta\ d\theta=2\int_0^{\Large\frac\pi2}\sin^5\theta\cos^2\theta\ d\theta, $$ then $$ \int_0^{\large\pi}\sin^5\theta\cos^2\theta\ d\theta=\frac{\Gamma\left(3\right)\cdot\Gamma\left(\dfrac32\right)}{\Gamma\left(\dfrac92\right)}=\frac{2!\cdot\Gamma\left(\dfrac32\right)}{\dfrac72\cdot\dfrac52\cdot\dfrac32\cdot\Gamma\left(\dfrac32\right)}=\large\color{blue}{\frac{16}{105}}, $$ where $\Gamma(n+1)=n\cdot\Gamma(n)$.


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