Integration : Problem 2

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

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