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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals3 Marks
Question
Evaluate the following integrals:$\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
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Answer

Let $\text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sqrt{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Bigg[\frac{2\text{t}^{\frac{3}{2}}}{3}\Bigg]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{\pi}{4}\Big)^{\frac{3}{2}}$
$\Rightarrow\text{I}=\frac{1}{12}\pi^{\frac{3}{2}}$

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