# What is $\int_0^{\pi/2}\sin^7(\theta)\cos^5(\theta)d\theta$

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

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

• If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly. Jan 6, 2019 at 18:09
• Possible duplicate Integral $\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx$
– A.Γ.
Jan 6, 2019 at 18:15
• $$\frac 12 B(4,3)$$? Jan 7, 2019 at 2:45

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 $$\dfrac 1{120}$$

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

• I think this is overkill but nevertheless, it's a solution too! Jan 6, 2019 at 21:23

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.

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 $$I = \frac12 \int_0^1u^5(1-u^2)^2\,du = \frac1{120}$$

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 $$I(7,5)=\frac{\Gamma(4)\Gamma(3)}{2\Gamma(7)}$$ $$I(7,5)=\frac1{120}$$