0
$\begingroup$

Evaluation of $\displaystyle \int\frac{1}{(a^2-\tan^2 x)\sqrt{b^2-\tan^2 x}}dx\;,$ Where $a>b$

$\bf{My\; Try::}$ Let $b^2-\tan^2 x= t^2\;,$ Then $-2\tan x \sec^2 xdx = 2tdt$

So integral convert into $\displaystyle I = -\int\frac{1}{(a^2+b^2-t^2)\cdot t}\times \frac{t}{(1+b^2-t^2)}\times \frac{1}{\sqrt{b^2-t^2}}dt$

Now how can i solve it, Help required, Thanks

$\endgroup$
  • 1
    $\begingroup$ Mathematica gives a truly disgusting answer, but it gives an answer, so have hope! It can be solved! $\endgroup$ – SAWblade Jan 26 '17 at 5:51
  • $\begingroup$ As @SAWblade said the answer is disgusting. Where did the question come from? Mathematica answer is: $$\frac{\sec (x) \sqrt{\left(b^2+1\right) \cos (2 x)+b^2-1} \left(\frac{\tanh ^{-1}\left(\frac{a \csc (x) \sqrt{\left(b^2+1\right) \cos (2 x)+b^2-1}}{\sqrt{2} \sqrt{b^2-a^2}}\right)}{a \sqrt{b^2-a^2}}-\frac{\tan ^{-1}\left(\frac{\csc (x) \sqrt{\left(b^2+1\right) \cos (2 x)+b^2-1}}{\sqrt{2} \sqrt{b^2+1}}\right)}{\sqrt{b^2+1}}\right)}{\sqrt{2} \left(a^2+1\right) \sqrt{b^2-\tan ^2(x)}}$$ $\endgroup$ – Ian Miller Jan 26 '17 at 5:59
3
$\begingroup$

I have an idea.

Substitute $u=\tan x $ to get $$I =\int \frac {1}{\sqrt {b^2-u^2}(u^2+1)(a^2-u^2)} \mathrm {d}u$$ Now substituting $u = b\sin v $, we get, $$I = \int \frac {1}{(b^2\sin ^2 v+1)(a^2-b^2\sin^2 v)}\mathrm {d}v $$ We then substitute for $w =\tan v $, to get, $$I =\int \frac {w^2+1}{((a^2-b^2)w^2+a^2)((b^2+1)w^2+1)} \mathrm {d}w $$ Now we do a partial fraction decomposition which I am unable to get right. Hope you can take it from here.

$\endgroup$
  • $\begingroup$ Just for fun : all of the above means that the direct change of variable could have been $x=\tan ^{-1}\left(\frac{b w}{\sqrt{w^2+1}}\right)$. NIce solution, indeed. $\endgroup$ – Claude Leibovici Jan 26 '17 at 9:30
  • $\begingroup$ Yes you are correct, @ClaudeLeibovici. But I could not have realised that in the first place. (+1). $\endgroup$ – Rohan Jan 26 '17 at 9:32
  • $\begingroup$ It is easy to see it after you gave all steps ! Cheers. $\endgroup$ – Claude Leibovici Jan 26 '17 at 9:55
3
$\begingroup$

Starting from $$I= \int\frac{dx}{(a^2-\tan^2 x)\sqrt{b^2-\tan^2 x}}$$ and using, as you did, $$b^2-\tan^2 x= t^2\implies x=-\tan ^{-1}\left(\sqrt{b^2-t^2}\right)\implies dx=\frac{t}{\sqrt{b^2-t^2} \left(b^2-t^2+1\right)}$$ $$I=\int\frac{dx}{\sqrt{b^2-t^2} \left(b^2-t^2+1\right) \left(a^2-b^2+t^2\right)}$$ The integrand can be decomposed as $$\frac{1}{a^2 \sqrt{b^2-t^2}}+\frac{\sqrt{b^2-t^2}}{\left(-a^2-1\right) \left(b^2-t^2+1\right)}-\frac{\sqrt{b^2-t^2}}{a^2 \left(a^2+1\right) \left(-a^2+b^2-t^2\right)}$$ leading, as given in comments, to $$a(a^2+1)\,I=\frac{\tan ^{-1}\left(\frac{a t}{\sqrt{a^2-b^2} \sqrt{b^2-t^2}}\right)}{\sqrt{a^2-b^2}}+\frac{a \tan ^{-1}\left(\frac{t}{\sqrt{b^2+1} \sqrt{b^2-t^2}}\right)}{\sqrt{b^2+1}}$$

Edit

Continuing from Rohan's answer, partial fraction decomposition leads, for the integrand, to $$\frac 1 {a^2+1}\left(\frac 1 {a^2+(a^2-b^2)w^2}+\frac 1{1+(b^2+1)w^2} \right)$$ which is much easier to integrate. This would lead to $$(a^2+1) \,I=\frac{\tan ^{-1}\left(\frac{ \sqrt{a^2-b^2}}{a}w\right)}{a \sqrt{a^2-b^2}}+\frac{\tan ^{-1}\left(\sqrt{b^2+1} w\right)}{\sqrt{b^2+1}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.