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Definite Integration

Section: Mathematics

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Question : 20 of 54

Marks: +1, -0

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

[AP EAPCET 21 May 2025 S2]

$\log (\sqrt{3}+1)$
$\log (\sqrt{3}-1)$
$\log (3+\sqrt{3})$
$\log (3-\sqrt{3})$

Solution:

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

\;I=\int\limits_{\pi/6}^{\pi/4} \frac{\cos\theta\,d\theta}{(\sin\theta+\cos\theta)\cos^2\theta}

Put

x=\sin\theta \Rightarrow dx=\cos\theta\,d\theta

=\int\limits_{\pi/6}^{\pi/4} \frac{d\theta}{\sin\theta\cos\theta+\cos^2\theta}

Dividing

N^T

and

D^r

by

\cos^2\theta

, we get

=\int\limits_{\pi/6}^{\pi/4} \frac{\sec^2\theta}{\tan\theta+1} d\theta

Put

\tan\theta+1=t,\ \sec^2\theta\,d\theta=dt

\;=\int_{1/\sqrt{3}+1}^{2} \frac{dt}{t}=[\ln t]_{1+1/\sqrt{3}}^{2}

\;=\ln 2-\ln (1+1/\sqrt{3})

\;=\ln 2-\ln \left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)=\ln \left(\frac{2\sqrt{3}}{\sqrt{3}+1}\right)

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

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