Question

Transform the integral
${\int }_1^2\frac{1}{x\sqrt{4+x^2}}dx$
by means of the substitution $x=\frac{1}{u}$. Hence, by means of further substitution, or otherwise, evaluate the integral.

Answer 1

Formula:
1. ${\tan }^2\theta +1={\sec }^2\theta$

2. $\int \sec \theta d\theta =\ln |\sec \theta +\tan \theta |$

Given,

${\int }_1^2\frac{1}{x\sqrt{4+x^2}}dx$

Let $x=\frac{1}{u}$

Then, $dx=-\frac{1}{u^2}du$

Upper limit: $u=\frac{1}{2}=0.5$

Lower limit: $u=1$

Substituting these values in the above integral we get,

${\int }_1^{0.5}\frac{-1}{\frac{1}{u}\sqrt{4+\frac{1}{u^2}}}.\frac{1}{u^2}du$

$\Rightarrow {\int }_1^{0.5}\frac{-u^2}{\sqrt{4u^2+1}}.\frac{1}{u^2}du$

$\Rightarrow {\int }_1^{0.5}\frac{-du}{\sqrt{4u^2+1}}$

Let $2u=tan\theta$

Then, $du=\frac{1}{2}{\sec }^2\theta$

Now, the integral becomes:

Let $I=\int \frac{-\frac{1}{2}{\sec }^2\theta }{\sqrt{{\tan }^2\theta +1}}d\theta$

$\Rightarrow I=\int \frac{-\frac{1}{2}{\sec }^2\theta }{\sec \theta }d\theta$

$\Rightarrow I=\int -\frac{1}{2}\sec \theta$

Using Formula 2, we get

$I=-\frac{1}{2}\ln |\sec \theta +\tan \theta |$

Since, $\tan \theta =2u$

$\sec \theta =\sqrt{1+{\tan }^2\theta }=\sqrt{1+4u^2}$

therefore, $I=-\frac{1}{2}\ln |2u+\sqrt{4u^1+1}|$

Substituting the upper and lower limits of $u$, we obatin

$I_1^{0.5}=-\frac{1}{2}|ln(2u+\sqrt{4u^2+1}){|}_1^{0.5}$

$=-\frac{1}{2}[ln(1+\sqrt{2})-\ln (2+\sqrt{5})]$

$=\frac{1}{2}[\ln (2+\sqrt{5})-\ln (1+\sqrt{2})]$

$=\frac{1}{2}\ln \left(\frac{2+\sqrt{5}}{1+\sqrt{2}}\right)$