19
$\begingroup$

$$\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx$$

My try:: $\displaystyle \int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx = \int_{1}^{2}\frac{\tan^{-1}x}{\tan^{-1}(x-1)-\tan^{-1}(x-2)}dx$

Now How can i solve after that.

plz help me

Thanks

$\endgroup$
  • $\begingroup$ It seems that there is no closed form solution. WolframAlpha gives the numerical result 1.10821.. $\endgroup$ – Erdos Yi Jun 9 '13 at 9:24
  • $\begingroup$ I think there's no close solution too. You may try numerical integration or any other different approach. $\endgroup$ – Dmoreno Jan 22 '14 at 9:52
3
$\begingroup$

After a few simple algebraic manipulations, the original integral turns into $$ \mathcal{I}=\frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{2}-\arctan\left(\frac{5-x^2}{12}\right)}{\frac{\pi}{2}+\arctan\left(\frac{3+x^2}{12}\right)}\,dx$$ which is pretty simple to approximate numerically, for instance through the Shafer-Fink approximation $\arctan(x)\approx\frac{3x}{1+2\sqrt{1+x^2}}$, leading to $\mathcal{I}\approx 1.108$. On the other hand, a simple closed form seems to be out of reach, since there is no evident symmetry and the substitution $\frac{3+x^2}{12}\mapsto\tan u$ does not simplify the integrand function as one might hope.

$\endgroup$
  • $\begingroup$ I have a problem (not knowing where). If I compute $\mathcal{I}$ as given in your answer, I end with $\mathcal{I}\approx 0.325805$. I must confess that I was unable to perform the few simple algebraic manipulations. $\endgroup$ – Claude Leibovici Dec 22 '17 at 7:28
1
$\begingroup$

Observe that the quadratic $ x^2 -3x +3 $ has discriminant $$b^2-4ac=(-3)^2-4(1)(3)=9-12=-3$$ Hence no real solutions, which means it never crosses the $x$-axis, and since it opens up, this means it is always positive. Using this, and the following formula for $\arctan \frac{1}{u}$ $$\arctan \frac{1}{u}=\frac{1}{2}\pi - \arctan u, u>0 $$ We can rewrite the denominator of our integral as $$\frac{1}{2}\pi - \arctan (x^2-3x+3)$$ having used $u=x^2-3x+3$ to apply the formula. We now have $$\int_1^2 \frac{\arctan x}{\frac{1}{2} \pi - \arctan (x^2-3x+3)}dx$$ This is as far as I got, and wolfram alpha didn't find a closed-form solution for that.

$\endgroup$

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.