First slide

Evaluation of definite integrals

Question

If $\int \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} = F (x)$ and $F (1) = 0$
then for $x > 0, F (x$ ) is equal to

Moderate

A

$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{2} x}{\sqrt{1 + x^{2}}}) + \frac{π}{\sqrt{2}}$
B

$\frac{1}{2} \tan^{- 1} (\frac{\sqrt{2} x}{\sqrt{1 + x^{2}}}) - \frac{π}{2 \sqrt{2}}$
C

$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{2 x}}{\sqrt{1 - x^{2}}}) + \frac{π}{2 \sqrt{2}}$
D

none of these

Solution

Putting $x = \frac{1}{t}$ so that $d x = - \frac{1}{t^{2}} d t$
We have
$\begin{array}{l} \int \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} = \int \frac{- t d t}{(1 + t^{2}) \sqrt{t^{2} - 1}} \\ = - \int \frac{d x}{2 + u^{2}} (t^{2} - 1 = u^{2}) \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} \frac{u}{2} + C \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} \frac{\sqrt{1 - x^{2}}}{x}) + C \end{array}$
Since $F (1) = 0, s o C = 0$ . Hence
$F (x) = - \frac{1}{\sqrt{2}} [\frac{π}{2} - \cot^{- 1} (\frac{1}{\sqrt{2}} \frac{\sqrt{1 - x^{2}}}{x})]$
$[\tan^{- 1} θ + \cot^{- 1} θ = π / 2]$
$= - \frac{π}{2 \sqrt{2}} + \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{2} x}{\sqrt{1 - x^{2}}})$
$(\begin{array}{l} \cot^{- 1} θ = \tan^{- 1} \frac{1}{θ}, for θ > 0 and \\ \cot^{- 1} θ = \tan^{- 1} \frac{1}{θ} + π, for θ < 0 \end{array})$

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