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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^{1 / 2} \frac{d x}{\left(1+x^2\right) \sqrt{1-x^2}}$ is equal to
  • ✓
    $\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{\frac{2}{3}}$
  • B
    $\frac{2}{\sqrt{2}} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)$
  • C
    $\frac{\sqrt{2}}{2} \tan ^{-1}\left(\frac{3}{2}\right)$
  • D
    $\frac{\sqrt{2}}{2} \tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
✓

Answer

Correct option: A.
$\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{\frac{2}{3}}$
(A)
Let $I =\int_0^{1 / 2} \frac{d x}{\left(1+x^2\right) \sqrt{1-x^2}}$
Put $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$
$\therefore \quad I=\int_0^{\pi / 6} \frac{\cos \theta d \theta}{\left(1+\sin ^2 \theta\right) \cos \theta}$
$\begin{array}{l}=\int_0^{\pi / 6} \frac{\sec ^2 \theta}{\sec ^2 \theta+\tan ^2 \theta} d \theta \\ =\int_0^{\pi / 6} \frac{\sec ^2 \theta}{1+2 \tan ^2 \theta} d \theta\end{array}$
$\begin{array}{l}=\int_0^{\pi / 6} \frac{\sec ^2 \theta}{1+(\sqrt{2} \tan \theta)^2} d \theta \\ =\frac{1}{\sqrt{2}} \int_0^{\pi / 6} \frac{\sqrt{2} \sec ^2 \theta}{1+(\sqrt{2} \tan \theta)^2} d \theta\end{array}$
$\begin{array}{l}=\frac{1}{\sqrt{2}}\left[\tan ^{-1}(\sqrt{2} \tan \theta)\right]_0^{\pi / 6} \\ =\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{\frac{2}{3}}\end{array}$

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