3
$\begingroup$

How to evaluate $$\int \frac{\arctan x}{x\sqrt {1+x^2}}e^{-\arctan x/x} \, dx$$

I tried using substitution $x=\tan t,dx=\sec ^2tdt,$It becomes the following: $$\int\frac{t}{\sin t}e^{-t\cot t}dt$$

Then I got stuck,any hint of help is welcome.

$\endgroup$
  • $\begingroup$ i think there is no solution in the known elementary functions $\endgroup$ – Dr. Sonnhard Graubner Sep 3 '17 at 11:51
  • $\begingroup$ Thank you.Isn't there a elementary function? $\endgroup$ – JamesJ Sep 3 '17 at 12:03
  • $\begingroup$ i think no try Wolfram Alpha $\endgroup$ – Dr. Sonnhard Graubner Sep 3 '17 at 12:03
5
$\begingroup$

We guess a primitive of the form $$\int\frac{t}{\sin t}e^{-t\cot t} dt = e^{-t\cot t}u$$ where $u$ is a function of $t$. Then $$\begin{aligned} &t\csc t{e^{ - t\cot t}} = ({e^{ - t\cot t}}u)' = (t{\csc ^2}t - \cot t){e^{ - t\cot t}}u + {e^{ - t\cot t}}u'\\ &\iff t\csc t = (t{\csc ^2}t - \cot t)u + u'\\ &\iff t\sin t = (t - \cos t\sin t)u + u'{\sin ^2}t \end{aligned}$$

One immediately notices that $u=\sin t $ is a solution.

Hence $$\int {\frac{t}{{\sin t}}{e^{ - t\cot t}}dt} = {e^{ - t\cot t}}\sin t + C$$

$\endgroup$

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.