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I have to integrate the following:

$\int_0^\limits\frac{\pi}{2}\sin^7(\theta)\cos^5(\theta)d\theta$

I decided to use a $u$ substitution of $u=\sin^2(\theta)$, and $\frac{du}{2}=\sin(\theta)\cos(\theta)$

and arrived at this integral

$\int_\limits{0}^{1}u^3(1-u)^2du$

From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$

I get the following:

$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\int_\limits{0}^{1}(u^2*(1-u)^3)du$$

Repeated again $g=u^2$, and $dv=(1-u)^3du$

$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\int_\limits{0}^{1}u(1-u)^4$$

Repeating again $g=u$, and $dv=(1-u)^4$

$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\biggl[\frac{u(1-u)^5}{5}\biggr]_0^1-\frac{1}5\int_\limits{0}^{1}(1-u)^5$$

and I get

$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\biggl[\frac{u(1-u)^5}{5}\biggr]_0^1-\frac{1}{30}\biggl[(1-u)^6\biggr]_0^1$$

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5 Answers 5

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I think you complicated the last part, after all you are integrating a polynomial.

$\displaystyle \int_0^1 u^3(1-u)^2\mathop{du}=\int_0^1 (u^3-2u^4+u^5)\mathop{du}=\left[\frac{u^4}4-2\frac{u^5}5+\frac{u^6}6\right]_0^1=\frac 14-\frac 25+\frac 16=\frac 1{60}$

Also you dropped the coeff $\dfrac 12$ from $\dfrac{du}2$, the result should be $\dfrac 1{120}$

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Note that: $$B(m+1,n+1)=2\int_0^{\pi/2}\cos^{2m+1}(\theta)\sin^{2n+1}(\theta)d\theta=\frac{m!n!}{(m+n+1)!}$$

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    $\begingroup$ I think this is overkill but nevertheless, it's a solution too! $\endgroup$
    – Frank W
    Jan 6, 2019 at 21:23
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I would just do $u=\sin\theta$ and $\mathrm du=\cos\theta\,\mathrm d\theta$. So\begin{align}\int_0^{\frac\pi2}\sin^7(\theta)\cos^5(\theta)\,\mathrm d\theta&=\int_0^{\frac\pi2}\sin^7(\theta)\bigl(1-\sin^2(\theta)\bigr)^2\cos(\theta)\,\mathrm d\theta\\&=\int_0^1u^7(1-u^2)^2\,\mathrm du.\end{align}I think that it's simpler.

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To lower the exponents a bit, notice that substitution $\theta \mapsto \frac\pi2-\theta$ yields $$\int_0^{\frac\pi2} \cos^7\theta\sin^5\theta\,d\theta = \int_0^{\frac\pi2} \sin^7\theta\cos^5\theta\,d\theta$$

so we have $$2I = \int_0^{\frac\pi2}\sin^5\theta\cos^5\theta(\cos^2\theta+\sin^2\theta)\,d\theta = \int_0^{\frac\pi2}\sin^5\theta\cos^5\theta\,d\theta = \int_0^{\frac\pi2}\sin^5\theta(1-\sin^2\theta)^2\cos\theta\,d\theta$$ Now setting $u = \sin\theta$ yields $$I = \frac12 \int_0^1u^5(1-u^2)^2\,du = \frac1{120}$$

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As has been noted, this integral can be related to the Beta Function. Here's how.

Consider the integral $$I(a,b)=\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx$$ If we make the substitution $t=\sin(x)^2$, we have that $$I(a,b)=\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt$$ $$I(a,b)=\frac12\int_0^1t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}\mathrm dt$$ We then recall the definition of the Beta function $$\mathrm B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\mathrm dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ Where $\Gamma(s)$ is the Gamma function. Technically it is defined by $$\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}\mathrm dx,\qquad \mathrm{Re}(s)>0$$ But in the case that $s$ is a (positive) integer, $$\Gamma(s)=(s-1)!$$ So without further adieu, $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ We then see that your integral is $$I(7,5)=\frac{\Gamma(4)\Gamma(3)}{2\Gamma(7)}$$ $$I(7,5)=\frac1{120}$$

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