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 $$\bowtie$$

Integration : Problem 2

$$\text{I have been inactive for quite a long time.}\\ \text{As usual, I will start this blogging streak}\\ \text{with a small warm-up exercise.}\\ \text{I will solve some mildly complex Integrals, }$$

$\underline{\text{Question 1.}} \\$ $\LARGE \int _{0}^{\frac{1}{2}}\frac{x^2}{(1+x^2)^{3/2}} \mathrm{d}x$
$\underline{\text{Solution}}$:
$$\text{Let, } I_{1} = \int _{0}^{\frac{1}{2}}\frac{x^2}{(1+x^2)^{3/2}} \mathrm{d}x \\ \text{We can write, } I_1 = \int _{0}^{\frac{1}{2}} \frac{x^2 + 1 - 1}{(1+x^2)^{3/2}} \mathrm{d}x \\ \implies I_1 = \int _{0}^{\frac{1}{2}} \frac{x^2+1}{(1+x^2)^{3/2}}\mathrm{d}x - \int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{Which with some algebraic manipulation comes to, }\\ I_1 = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1+x^2}}\mathrm{d}x - \int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{Let,} I_2 = \int \frac{1}{\sqrt{1+x^2}}\mathrm{d}x \\ \text{and, Let , } I_3 = \int \frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{We know that, } \boxed{\boxed{\int \frac{\mathrm{d}x}{\sqrt{x^2 + a^2}} = \log|x +\sqrt{x^2 + a^2}|+c }} \\ \therefore I_2 = \int \frac{1}{\sqrt{1+x^2}}\mathrm{d}x = \log|x+\sqrt{1+x^2}| \\ \text{ neglecting } c \text{ because it will ultimately} \\ \text{ be a definite integral }.\\ \\ \text{Now the rigor starts, } I_3 = \int \frac{1}{(1+x^2)^{3/2}}\mathrm{d}x\\ \text{Take, } x = \tan {t} \rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = \sec^2 {t} \\ \therefore I_3 = \int \frac{\sec^2{t}}{(1+\tan^2{t})^{3/2}} \mathrm{d}t\\ \text{We know, } \boxed{\boxed{1 + \tan^2 \theta = \sec^2 \theta}} \\ \text{Using this } I_3 \text{ changes to, }\\ I_3 = \int \frac{1}{\sec t} \mathrm{d} t \implies \int \cos {t} \mathrm{d}t = \sin t \\ \text{We get the substitution in place ,}\\ t = \tan^{-1} x \rightarrow \sin t \rightarrow \sin{(\tan^{-1} x)} = \frac{x}{\sqrt{x^2+1}} \\ \therefore I_3 = \sin t = \frac{x}{\sqrt{x^2+1}} \text{Now,} \\ I_1 = I_2 - I_3 \text{ with no limits. }\\ \therefore I_1 = \int \frac{1}{\sqrt{1+x^2}}\mathrm{d}x = \log|x+\sqrt{1+x^2}| - \frac{x}{\sqrt{x^2+1}}\\ \text{Put limits } \frac{1}{2} \text{ to } 0 \text{ and get answer as, }\\ I_1 = {I_1}|_{0}^{1/2}=\boxed{0.033}.\\ \text{Cheers! } \bowtie$$