A Nice Problem Involving Gamma and Beta Function.

[Playing with Gamma and Beta functions].
Solve :
$$\int_{0}^{\pi/2}(\cos^{2}x-\cos^{3}x) \mathrm{d}x$$

Let,
$$I = \int_{0}^{\pi/2}(\cos^{2}x-\cos^{3}x) \mathrm{d}x\tag*{(1)}$$

Beta function is defined as:
$$\boxed{B(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{(y-1)}\mathrm{d}t \hspace{0.2cm}\forall x, y >0}$$

Now, there are many properties of this beautiful mathematical tool, some of them are,

1. $$B(x,y) = B(y,x)$$ They are symmetric.
2. We can write Beta Functions in terms of Gamma Functions as follows, $$B(x,y) = \frac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}.$$
3. We can obviously add factorial into the mix too, $$B(x,y)=\frac{(x-1)!(y-1)!}{(x+y-1)}$$
4. We can define it in terms of trigonometric angles too,
   $$B(x,y)=2 \int_{0}^{\pi/2}(\sin \theta)^{2x-1}(\cos \theta)^{2y-1}\mathrm{d}\theta$$

Now,
$$I = \int_{0}^{\infty}(\cos^{2}x)\mathrm{d}x-\int_{0}^{\infty}(\cos^{3}x) \mathrm{d}x \\ \implies I = I_1 + I_2$$

We will solve both separately,
$$I_1 = \int_{0}^{\infty}(\cos^{2}x)\mathrm{d}x \\ \implies I_1 = \int_{0}^{\infty}(\sin x)^0(\cos x)^2\mathrm{d}x \\ \implies I_1 = \int_{0}^{\infty} (\sin x)^{2.\frac{1}{2}-1}(\cos x)^{2.\frac{3}{2}-1}\mathrm{d}x \\ \implies I_1 = \frac{B(\frac{1}{2},\frac{3}{2})}{2}\\ \implies \boxed{ I_1 = \frac{\Gamma(\frac{1}{2}). \Gamma(\frac{3}{2})}{2\Gamma(\frac{1}{2}+\frac{3}{2})}}$$
Similarly, for the other integral,
$$I_2 = \int_{0}^{\infty}(\cos^{3}x)\mathrm{d}x\\ \implies I_2 = \int_{0}^{\infty}(\sin x)^0 (\cos x)^3 \mathrm{d}x\\ \implies I_2 = \int_{0}^{\infty}(\sin x)^{2.\frac{1}{2}-1}(\cos x)^{2.2-1}\mathrm{d}x\\ \implies I_2 = \frac{B(\frac{1}{2},2)}{2}\\ \implies \boxed{ I_2 = \frac{\Gamma(\frac{1}{2}). \Gamma(2)}{2\Gamma(\frac{1}{2}+2)}}$$
Therefore,
$$\boxed{\int_{0}^{\infty}(\cos^{2}x-\cos^{3}x) \mathrm{d}x = \frac{\Gamma(\frac{1}{2}). \Gamma(\frac{3}{2})}{2\Gamma(\frac{1}{2}+\frac{3}{2})}+\frac{\Gamma(\frac{1}{2}). \Gamma(2)}{2\Gamma(\frac{1}{2}+2)}}$$
Cheers $$\bowtie$$