₀¹/2} {dx}{(1+X²){1-X²} is equal to

Correct Answer is : 12{1}{{2}}21 tan⁡−123^{-1}{2/3}tan−132

Correct Answer is : 12{1}{{2}}21 tan⁡−123^{-1}{2/3}tan−132

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KCET 2018 Math Solved Paper

Question : 2 of 60

Marks: +1, -0

\int\limits_{0}^{1/2} \frac{dx}{(1+X^2)\sqrt{1-X^2}}

is equal to

Solution:

Given that,

\int\limits_{0}^{1/2} \;\frac{dx}{(1+x^{2})\sqrt{1-x^{2}}}

Let

x=\sin\theta

then

dx=\cos\theta d\theta

and limits become

0\rightarrow 0,\;1/2\rightarrow \pi/6

.So,

\int\limits_{0}^{\pi/6} \;\frac{\cos\theta\,d\theta}{(1+\sin^{2}\theta)\sqrt{1-\sin^{2}\theta}}

=\int\limits_{0}^{\pi/6} \;\frac{\cos\theta\,d\theta}{(1+\sin^{2}\theta)\cos\theta}

=\int\limits_{0}^{\pi/6} \;\frac{\sec^{2}\theta}{1+2\tan^{2}\theta}\,d\theta

Let

u=\sqrt{2}\tan\theta

then

du=\sqrt{2}\sec^{2}\theta\,d\theta

and limits becomes

0\rightarrow 0,\;\frac{\pi}{6}\rightarrow\sqrt{\frac{2}{3}}

.So,

\;\frac{1}{\sqrt{2}} \int\limits_{0}^{\sqrt{\frac{2}{3}}} \;\frac{1}{1+u^{2}}\,du

We know that

\int\;\frac{1}{1+x^{2}}\,dx = \tan^{-1} x

.So,

\frac{1}{\sqrt{2}} \tan^{-1} u \Big|_{0}^{\sqrt{2/3}} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) - 0

=\;\frac{1}{\sqrt{2}} \tan^{-1}\left(\sqrt{\frac{2}{3}}\right)

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