How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $

Question

How to solve $$\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $$

I have tried substitution of cosine and sine, dividing and multiplying by Sine and cosine, converting everything to tan. It can probably be solved using beta/gamma function but I can't figure it out.

Answer 1

Let $\theta =2t $ $$I=\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4d\theta = 128\int_0^{\pi/2} (\cos^{10}t - \cos^{12} t )dt$$ Then, apply the recursive formula \begin{align} K_n= \int_0^{\pi/2} \cos^ntdt = \frac{n-1}n K_{n-2},\>\>\>\>\>K_0=\frac\pi2 \end{align} to arrive at

$$I=128 (K_{10}- K_{12} )=\frac{32}3K_{10}=\frac{32}3\cdot\frac9{10}\frac78\frac56\frac34\frac12K_0=\frac{21\pi}{16}$$

Answer 2

The integral is equal to $$\frac12\int_{-\pi}^\pi\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^2\left(1+\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^4d\theta=-\frac{1}{128}\int_{-\pi}^\pi e^{-6i\theta}(e^{2i\theta}-1)^2(e^{i\theta}+1)^8\,d\theta,$$ and if we imagine the integrand written as a linear combination of exponentials, then all the exponentials vanish after the integration, and only its constant term "survives". It is equal to $$[z^6](z^2-1)^2(z+1)^8=\binom{8}{2}-2\binom{8}{4}+\binom{8}{6}=28-2\cdot70+28=-84,$$ so the integral is equal to $\dfrac{84}{128}\cdot2\pi=\dfrac{21}{16}\pi$.

Answer 3

Here is an answer using just normal methods. We have:$$\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta$$

Now let's simplify the trigonometric expression. Let $c=\cos\theta$:

$$\begin{align}\sin ^2 \theta(1 + \cos \theta)^4=(1-c^2)(1+c)^4=(1-c^2)(c^4+4c^3+6c^2+4c+1)&\\= -c^6-4c^5-5c^4+5c^2+4c+1\end{align}$$

Starting from $c^6$:

$$\cos^6\theta=\dfrac18(1+\cos2\theta)^3=\dfrac18(1+3\cos2\theta+\dfrac32(1+\cos4\theta)+\cos^32\theta)$$

Note that, the antiderivatives of $\cos (m\theta)$ and $\cos^{2k+1}(m\theta)$ have only the sine function in their argument and since we are integrating from $0$ to $\pi$ the integrals are zero. So, for $\cos^6\theta$ we are only left with:

$$-\int_0^{\pi}\cos^6\theta=-\int_0^{\pi}(\dfrac18+\dfrac3{16}) d\theta=-\dfrac{5\pi}{16}$$

Similarly, for $-4c^5+4c$ the integral is zero. Now for $-5c^4+5c^2$ we have:

$$-5\cos^4\theta+5\cos^2\theta=5\cos^2\theta(1-\cos^2\theta)=\dfrac54\sin^22\theta=\dfrac58(1-\cos4\theta)$$So the integral is $\dfrac58\pi$. We are only left with the term $1$ which has $\pi$ as the integral. Altogether, the integral is equal to $-\dfrac{5\pi}{16}+\dfrac{5\pi}8+\pi=\dfrac{21\pi}{16}$

Answer 4

There is another solution using the "reverse" of multiple angle formulae. In your case $$(1-\cos^2(t))(1+\cos(t))^4=$$ $$\frac{3 \cos (t)}{2}-\frac{15 \cos (2 t)}{32}-\frac{5\cos (3 t)}{4} -\frac{13\cos (4 t)}{16} -\frac{\cos (5 t)}{4} -\frac{\cos (6 t)}{32} +\frac{21}{16}$$

Have a look here.

Answer 5

" It can probably be solved using beta/gamma function but I can't figure it out."

Well here's the beta function approach,

$$I=\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4\, d\theta$$

Use $\theta=2x,$ $$I=2\int_0^\frac{\pi}{2} \sin ^2 2x(1 + \cos 2x)^4 \,dx$$

$$I=2\int_0^\frac{\pi}{2} (4\sin^2 x \cos^2 x)(2\cos^2x)^4 dx\implies 128\int_0^\frac{\pi}{2} \sin^2 x \cos^{10} x\,dx$$

$$\frac{1}{128}I=\int_0^\frac{\pi}{2} \sin^2 x \cos^{10} x\,dx$$

Use the below formula,

$$\color{red}{\frac{1}{2}\beta(p, q) = \int_0^{\frac{\pi}{2}} \sin^{2p-1}x \, \cos^{2q-1}x \, dx}$$

On comparing,

$$p=\frac{3}{2}, q=\frac{11}{2}$$

$$\frac{1}{128}I=\frac{1}{2}\beta\left(\frac{3}{2}, \frac{11}{2}\right)$$

$$\boxed{I=64\beta\left(\frac{3}{2}, \frac{11}{2}\right)}$$

We have utilised one of the integral forms from the definition of beta function,

$$\boxed{I=\frac{21\pi}{16}}$$

Answer 6

Idea: use $\sin^2\theta = 1- \cos^2 \theta$ and work out the polynomial in $\cos \theta$, and apply techniques from here, e.g., or here on this site in a more modern look.