Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 4

Year : 2013

$$Let \space I =\int\limits_{x=0}^{x=1}\int\limits_{y=0}^{y=\sqrt{1+x^2}}\dfrac 1{\sqrt{(1+x^2)^2}+y^2}dxdy$$

Integrating w.r.t. y we get

$I=\int\limits_{x=0}^{x=1}\Bigg[\dfrac 1{\sqrt{1+x^2}}\tan^{-1}\dfrac y{\sqrt{1+x^2}}\Bigg]_0^{\sqrt{1+x^2}} dx \\ =\int\limits_0^1\dfrac 1{\sqrt{1+x^2}}\Bigg[\tan^{-1}\Bigg(\dfrac {\sqrt{1+x^2}}{\sqrt{1+x^2}}\Bigg)-\tan^{-1}(0)\Bigg]dx\\ =\int \limits_0^1\dfrac 1{\sqrt{1+x^2}}\times \dfrac \pi4dx\\ =\dfrac \pi4\Big[\log |x+\sqrt{1+x^2}|\Big]_0^1\\ =\dfrac \pi4 \Big[\log|1+\sqrt2|-\log|0+\sqrt1|\Big]\\ =\dfrac \pi4 \log(1+\sqrt2)\\ \because \log(1)=0 $