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

Marks : 4

Year : 2014

Integrating first w.r.t. y treating x constant.

$ I= \int\limits_0^a\int\limits_0^{\sqrt{1+x^2}}\dfrac 1{\sqrt{(1+x^2)+y^2}}dxdy \\ I=\int\limits_0^a\Bigg[\dfrac 1{\sqrt{1+x^2}} \tan^{-1}\dfrac y{\sqrt{1+x^2}}\Bigg]_0^{\sqrt{1+x^2}}dx \\ I=\int\limits_0^a\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}}\Bigg(\dfrac \pi4\Bigg)dx \\ =\dfrac \pi4[\log|x+\sqrt{1+x^2}|]^a_0 \\ =\dfrac \pi4[\log(a+\sqrt{1+a^2})-\log 0+\sqrt 1] \\ =\dfrac \pi4\log(a+\sqrt{1+a^2})$