Trig integral with sine and cosine

Question

What sort of formulas can I use to reduce this into something I can work with?

$$3a^2\int_{0}^{2\pi} \sin^2(\theta)\cos^4(\theta) \, d\theta$$

Answer 1

You could use double-angle formulas. We have $\sin^2\theta\cos^2\theta=\frac{1}{4}\sin^2 2\theta$ and $\cos 2\theta=2\cos^2\theta-1$. Thus our function is $$\frac{1}{8}(\sin^2 2\theta \cos 2\theta+\sin^2 2\theta).$$

To integrate $\frac{1}{8}\sin^2 2\theta \cos 2\theta$, we can let $u=\sin 2\theta$. The definite integral will be $0$. It remains to find $$\int_0^{2\pi}\frac{1}{8}\sin^2 2\theta\,d\theta,$$ which is left for you to do.

Answer 2

Hint : \begin{align} 3a^2\int_{0}^{2\pi} \sin^2(\theta)\cos^4(\theta) \, d\theta&=3a^2\int_{0}^{2\pi} (1-\cos^2(\theta))\cos^4(\theta) \, d\theta\\ &=3a^2\int_{0}^{2\pi} \cos^4(\theta) \, d\theta-3a^2\int_{0}^{2\pi} \cos^6(\theta) \, d\theta\\ \end{align} Use integration by reduction formula and evaluate first $\displaystyle3a^2\int_{0}^{2\pi} \cos^6(\theta) \, d\theta$, then the result merges with $\displaystyle3a^2\int_{0}^{2\pi} \cos^4(\theta) \, d\theta$.

Answer 3

A less common way is to use Euler's formulae:

$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Your integrand becomes:

$$\bigg(\frac{e^{ix}-e^{-ix}}{2i}\bigg)^2\bigg(\frac{e^{ix}+e^{-ix}}{2}\bigg)^4$$

After doing the exponentations and the multiplications you're left with

$$\frac{1}{64}e^{-2ix}+\frac{1}{64}e^{2ix}-\frac{1}{32}e^{-4ix}-\frac{1}{32}e^{4ix}-\frac{1}{64}e^{-6ix}-\frac{1}{64}e^{6ix}+\frac{1}{16}$$

Now you can switch back to sines and cosines:

$$\frac{1}{32}(\cos{2x}-2\cos{4x}-\cos{6x}+2)$$

The last expression is very easy to integrate

Answer 4

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 $$ 3a^2\int_{0}^{\large2\pi} \sin^2\theta\cos^4\theta \, d\theta=12a^2\int_0^{\Large\frac\pi2}\sin^2\theta\cos^4\theta\ d\theta, $$ then $$ 3a^2\int_{0}^{\large2\pi} \sin^2\theta\cos^4\theta \, d\theta=6a^2\cdot\frac{\Gamma\left(\dfrac32\right)\cdot\Gamma\left(\dfrac52\right)}{\Gamma\left(4\right)}=6a^2\cdot\frac{\dfrac38\cdot\Gamma^2\left(\dfrac12\right)}{3!}=\large\color{blue}{\frac{3}{8}\pi a^2}, $$ where $\Gamma(n+1)=n\cdot\Gamma(n)$ and $\Gamma\left(\dfrac12\right)=\sqrt\pi$.