lf the value of the integral ∫01/2 x21−x23/2dx is k6, then k is equal to

A
$2 \sqrt{3} + π$
B
$2 \sqrt{3} - π$
C
$3 \sqrt{2} + π$
D
$3 \sqrt{2} - π$

Solution:

We have,

$\int_{0}^{1 / 2} \frac{x^{2}}{{(1 - x^{2})}^{3 / 2}} d x = \frac{k}{6}$

$\Rightarrow \int_{0}^{π / 6} \frac{\sin^{2} θ}{\cos^{3} θ} \cos θ d θ = \frac{k}{6}$ , where $x = \sin θ$

$\begin{array}{l} \Rightarrow \int_{0}^{π / 6} \tan^{2} θ d θ = \frac{k}{6} \\ \Rightarrow \int_{0}^{π / 6} (\sec^{2} θ - 1) d θ = \frac{k}{6} \\ \Rightarrow [\tan θ - θ]_{0}^{π / 6} = \frac{k}{6} \Rightarrow \frac{1}{\sqrt{3}} - \frac{π}{6} = \frac{k}{6} \Rightarrow k = 2 \sqrt{3} - π . \end{array}$

lf the value of the integral $\int_{0}^{1 / 2} \frac{x^{2}}{{(1 - x^{2})}^{3 / 2}} d x$ is $\frac{k}{6}$ , then $k$ is equal to

Solution:

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