Question
Evaluate the following integrals:
$\int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}$

Answer

Let $\text{a}^2+\text{x}^2=\text{t}^2$
Differentiating w.r.t. x, we get
$2\text{xdx}=2\text{tdt}$
$\text{xdx}=\text{tdt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\text{a}\Rightarrow\text{t}=\sqrt{2}\text{a}$
$\therefore\ \int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\frac{\text{t dt}}{\text{t}}$
$=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\text{dt}$
$=\big[\text{t}\big]^{\sqrt{2}\text{a}}_\text{a}$
$=\big[\sqrt{2}\text{a}-\text{a}\big]$
$=\text{a}\big(\sqrt{2}-1\big)$

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