Hello, satishinamdar!
Am I missing something?
. . There is no easy way to integrate this . . .
pka said:
\(\displaystyle \L\frac{1}{{x\,+\,\sqrt {a^2 - x^2 } }}\cdot\left({\frac{{x\,-\,\sqrt{a^2\,-\,x^2}}}{{x\,-\,\sqrt {a^2-x^2 }}}}\right)\:=\:\frac{-1}{{a^2 }}\left( {x\,-\,\sqrt{a^2 - x^2}} \right)\) . This is incorrect!
The denominator is: \(\displaystyle \
x\,+\,\sqrt{a^2\,-\,x^2})(x\,-\,\sqrt{a^2\,-\,x^2})\:=\:x^2\,-\,(a^2\,-\,x^2)\:=\:2x^2\,-\,a^2\)
The integral is far more difficult than you think . . .
Let \(\displaystyle x\,=\,a\cdot\sin\theta\;\;\Rightarrow\;\;dx\,=\,a\cdot\cos\theta\,d\theta\)
Note that: \(\displaystyle \sqrt{a^2\,-\,x^2}\:=\:\sqrt{a^2\,-\,a^2\sin^2\theta}\:=\:\sqrt{a^2(1\,-\,\sin^2\theta)}\:=\:\sqrt{a^2\cos^2\theta}\:=\:a\cdot\cos\theta\)
Substitute: \(\displaystyle \L\:\int\frac{a\cdot\cos\theta\,d\theta}{a\cdot\sin\theta\,+\,a\cdot\cos\theta}\;= \;\int\frac{\cos\theta}{\sin\theta\,+\,\cos\theta}\,d\theta\)
Divide top and bottom by \(\displaystyle \cos\theta:\;\;\L\int\frac{d\theta}{\tan\theta\,+\,1}\)
Let \(\displaystyle v\,=\,\tan\theta\;\;\Rightarrow\;\;\theta\,=\,\arctan v\;\;\Rightarrow\;\;d\theta\,=\,\frac{dv}{v^2\,+\,1}\)
Substitute: \(\displaystyle \L\:\int\frac{\left(\frac{dv}{v^2\,+\,1}\right)}{v\,+\,1}\;=\;\int\frac{dv}{(v\,+\,1)(v^2\,+\,1)}\)
Partial fractions: \(\displaystyle \L\:\frac{1}{(v\,+\,1)(v^2\,+\,1)}\;=\;\frac{A}{v\,+\,1}\,+\,\frac{Bv\,+\,C}{v^2\,+\,1}\)
. . and we get: \(\displaystyle \,A\,=\,\frac{1}{2},\:B\,=\,-\frac{1}{2},\:C\,=\,\frac{1}{2}\)
The integral becomes: \(\displaystyle \L\:\int\left(\frac{\frac{1}{2}}{v\,+\,1}\,+\,\frac{-\frac{1}{2}v}{v^2\,+\,1}\,+\,\frac{\frac{1}{2}}{v^2\,+\,1}\right)\,dv\)
. . . \(\displaystyle \L=\;\frac{1}{2}\left[\int\frac{dv}{v\,+\,1}\,-\,\int\frac{v\,dv}{v^2\,+\,1}\,+\,\int\frac{dv}{v^2\,+\,1}\right)\)
. . . \(\displaystyle \L=\;\frac{1}{2}\left[\ln|v\,+\,1|\,-\,\frac{1}{2}\ln(v^2\,+\,1)\,+\,\arctan v\right]\)
Multiply by \(\displaystyle \frac{2}{2}:\L\;\;\frac{1}{4}\left[2\cdot\ln|v\,+\,1|\,-\,\ln(v^2\,+\,1)\,+\,2\cdot\arctan v\right]\)
. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln(v\,+\,1)^2 - \ln(v^2\,+\,1) + 2\cdot\arctan v\right]\)
. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln\frac{(v\,+\,1)^2}{v^2\,+\,1}\,+\,2\cdot\arctan v\right]\)
Back-substitute: \(\displaystyle \L\:\frac{1}{4}\left[\ln\frac{(\tan\theta\,+\,1)^2}{\tan^2\theta\,+\,1}\,+\,2\cdot\arctan(\tan\theta)\right]\)
. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln\frac{(\tan\theta\,+\,1)^2}{\sec^2\theta}\,+\,2\theta\right]\)
Back-substitute: Since \(\displaystyle \sin\theta\,=\,\frac{x}{a}\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{x}{\sqrt{a^2-x^2}},\:\sec\theta\,=\,\frac{a}{\sqrt{a^2-x^2}}\)
And we have: \(\displaystyle \L\:\frac{1}{4}\left[\ln\frac{\left(\frac{x}{\sqrt{a^2-x^2}}\,+\,1\right)^2}{\left(\frac{a}{\sqrt{a^2-x^2}}\right)^2}\,+\,2\cdot\arcsin\left(\frac{x}{a}\right)\right]\)
. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln\frac{(x\,+\,\sqrt{a^2-x^2})^2}{a^2}\,+\,2\cdot\arcsin\left(\frac{x}{a}\right)\right]\)
. . . \(\displaystyle \L=\;\frac{1}{2}\left[\ln\frac{x\,+\,\sqrt{a^2-x^2}}{a}\,+\,\arcsin\left(\frac{x}{a}\right)\right]^a_0\)
We have: \(\displaystyle \L\:\frac{1}{2}\left[\ln\left(\frac{a\,+\,\sqrt{0}}{a}\right)\,+\,\arcsin\left(\frac{a}{a}\right)\right]\,-\,\frac{1}{2}\left[\ln\left(\frac{0\,+\,\sqrt{a^2}}{a}\right)\,+\,\arcsin\left(\frac{0}{a}\right)\right]\)
. . . \(\displaystyle \L=\;\frac{1}{2}\left[\ln(1)\,+\,\arcsin(1)\right]\,-\,\frac{1}{2}\left[\ln(1)\,+\,\arcsin0\right]\)
. . . \(\displaystyle \L=\;\frac{1}{2}\left[0\,+\,\frac{\pi}{2}\right]\,-\,\frac{1}{2}\left[0\,+\,0\right}\)
. . . \(\displaystyle \L=\;\frac{\pi}{4}\)
. . . . There!
