ISRO CSE 2016 | Question: EC Calculus

Question

Evaluate $\int_0^1 \int_0^{\sqrt{1+x^2}} \frac{d x \cdot d y}{\left(1+x^2+y^2\right)}$

• A. $\frac{\pi}{2}[\log (1+\sqrt{2})]$
• B. $\frac{\pi}{4}[\log (1+\sqrt{2})]$
• C. $\frac{\pi}{2}[\log (1-\sqrt{2})]$
• D. $\frac{\pi}{4}[\log (1-\sqrt{2})]$

Answer 1

$\int_{0}^{1} \ \int_{0}^{\sqrt{1 + x^2}} \ \dfrac{dx.dy}{1+x^2 + y^2}$

 

Let $\ (1 + x^2)$  be   $a$

 

$\int_{0}^{1} \ \int_{0}^{\sqrt{1 + x^2}} \ \dfrac{dx.dy}{(\sqrt{a})^2 + y^2}$

 

$\int_{0}^{1} \ dx. \int_{0}^{\sqrt{1 + x^2}} \ \dfrac{dy}{(\sqrt{a})^2 + y^2}$

 

$\int_{0}^{1} \ dx \ .[ \dfrac{1}{\sqrt{a}}.tan^{-1}\dfrac{y}{\sqrt{a}}]$ $\left.\begin{matrix} & \sqrt{1+x^2} \\ \ \ \ \ & 0 \end{matrix}\right|$                                                          $\because \int \dfrac{1}{a^2 + x^2} = \dfrac{1}{a}tan^{-1}\dfrac{x}{a}$

 

$\int_{0}^{1} \begin{Bmatrix} \dfrac{1}{\sqrt{1+x^2}}{tan}^{-1} \dfrac{\sqrt{1+x^2}}{\sqrt{1+x^2} } - \dfrac{1}{\sqrt{1+x^2}}{tan}^{-1} \dfrac{0}{\sqrt{1+x^2}} \end{Bmatrix}dx$

 

$\int_{0}^{1} \begin{Bmatrix} \dfrac{1}{\sqrt{1+x^2}}{tan}^{-1} 1 \end{Bmatrix}dx$                   $\because tan^{-1}1 = \dfrac{\pi }{4}$

 

$\int_{0}^{1} \dfrac{\pi }{4\sqrt{1+x^2}} dx$

 

$\dfrac{\pi }{4} log(x + \sqrt{x^2 + 1})\left.\begin{matrix} & 1 \\ \ & 0 \end{matrix}\right|$                   ${\color{Red} \because  \dfrac{1}{\sqrt{a^2 + x^2}} = log(x+\sqrt{a^2 + x^2})}$

 

$\dfrac{\pi }{4}\begin{Bmatrix} log(1 + \sqrt{1 + 1}) - log(0 + \sqrt{0 + 1}) \end{Bmatrix}$

 

$\dfrac{\pi }{4}\left \{ log(1 + \sqrt{2}) - log(1) \right \}$

 

$\dfrac{\pi }{4}log(1 + \sqrt{2})$