How can I evaluate the following integral? $$\int(\sqrt{x}-x)(e^{\arctan\sqrt{x}})^2dx$$

I'd like the whole solution if possible. I tried using the substitution: $\sqrt{x}=t$, followed by: $2\arctan{t}=m$, to get: $$\int e^m\tan^2{\frac{m}{2}}\sec^2{\frac{m}{2}}\left[1-\tan{\frac{m}{2}}\right]dm$$ But it doesn't get me anywhere.

A complete solution will be sincerely appreciated.

  • 1
    $\begingroup$ Maybe it helps to know that the result is $-\frac{1}{2}(\sqrt{x}-1)^2 (x+1) e^{2 \arctan\sqrt{x}}+C$ . $\endgroup$ – user90369 Feb 16 '17 at 11:33
  • $\begingroup$ I'd like to use a method without involving complex variables. Additionally, I know the result - my brain seems to have shut down unfortunately, despite the problem being so simple. :( $\endgroup$ – Kugelblitz Feb 16 '17 at 11:34
  • 1
    $\begingroup$ "despite the problem being so simple" ??? $\endgroup$ – Yves Daoust Feb 16 '17 at 11:35
  • $\begingroup$ Yea. In a worksheet of many integrals, this happens to be one of the easier ones as classified by section headings. $\endgroup$ – Kugelblitz Feb 16 '17 at 12:19
  • $\begingroup$ @Kugelblitz. Could you tell where they say that the problem is simple ? I am just curious. Cheers. $\endgroup$ – Claude Leibovici Feb 16 '17 at 14:03

Assume the integral can be written in the form $g(x) = f(x) e^{2\arctan \sqrt{x}}$ for some unknown function $f(x).$ Then $$ g'(x) = \left(f' + \frac{f}{\sqrt{x}(1+x)}\right)e^{2\arctan \sqrt{x}} $$ must be the integrand, meaning $$ f' + \frac{f}{\sqrt{x}(1+x)} = \sqrt{x}-x $$ or $$ \sqrt{x}(1+x)f' + f = x - x^{3/2} + x^2 - x^{5/2}. $$

Now find a particular solution of this ODE for $f$ via the method of undetermined coefficients with $f(x)=a_0 + a_1\sqrt{x} + a_2 x + a_3 x^{3/2} + a_4x^2.$ This results in six linear equations in the five unknown $a_i.$ The six equations are consistent though, with $a_4=-1/2,$ $a_3=1,$ $a_2=-1,$ $a_1=1,$ and $a_0=-1/2.$

  • $\begingroup$ Mathematica gives: $-\frac{1}{2} \left(\sqrt{x}-1\right)^2 (x+1) e^{2 \tan ^{-1}\left(\sqrt{x}\right)}$ $\endgroup$ – David G. Stork Feb 19 '17 at 23:19
  • $\begingroup$ @DavidG.Stork: I know (this is stated in the comments for the question and I also checked Wolfram Alpha). My answer is just the expansion of this product $\endgroup$ – J. Heller Feb 19 '17 at 23:38
  • $\begingroup$ I see. There must, however, be a more elegant method using integration techniques. Nevertheless, thank you. $\endgroup$ – Kugelblitz Feb 20 '17 at 2:56
  • $\begingroup$ Possibly there is some other way, but my answer is pretty easy. The system of equations for the $a_i$ is upper triangular with a lot of zeros in the upper triangle, so it's easy to solve by hand. $\endgroup$ – J. Heller Feb 20 '17 at 5:44

To solve such a problem without literature and programs I would set

$\displaystyle \int (\sqrt{x}-x)e^{2\arctan\sqrt{x}}dx = 2\int (t^2-t^3)e^{2\arctan t}dt := 2p(t)e^{2\arctan t} + C$

with $\enspace x=t^2$ and knowing that $\enspace \displaystyle (\arctan t)’=\frac{1}{1+t^2}$ . $\enspace p(t)$ is a polynom.

Such methods I’ve learned in the school (means: it's nothing special).

The derivation of both sides by $\enspace t\enspace $ and multiplicating by $\enspace 1+t^2\enspace $ gives

$\displaystyle (t^2-t^3)(1+t^2)=(1+t^2)p’(t)+p(t)\enspace $ and we know now that the degree of the left side is $\enspace 5$

and it follows for $\enspace p(t)\enspace $ the degree $\enspace 4$ : $\enspace p(t):=a+bt+ct^2+dt^3+et^4$ .

Comparing the coefficients we get $\enspace \displaystyle (a;b;c;d;e)=(-\frac{1}{4};\frac{1}{2};-\frac{1}{2};\frac{1}{2};-\frac{1}{4})$

and therefore $\enspace \displaystyle p(t)=-\frac{1}{4}(1-2t+2t^2-2t^3+t^4)\enspace $ with $\enspace t=\sqrt{x}$.

  • $\begingroup$ Where is the $2t dt$ term? I only see the $2$ without the $t$. What is your guess for $p(t)\exp(..)$ based on? What is the name of that integration technique ? I have never got that taught in my school. $\endgroup$ – Shashi Feb 22 '17 at 17:05
  • 2
    $\begingroup$ @Shashi : Thanks for the hint with $t$, it was only a writing mistake. For integration you learn a lot of different methods. I don't know all the names but here we have a reduction to the inhomogeneous linear differential equation of the first order. $\int a(x)e^{f(x)}dx=b(x)e^{f(x)}+C$ can be solved by $b'(x)+f'(x)b(x)=a(x)$ . $\endgroup$ – user90369 Feb 22 '17 at 21:24
  • $\begingroup$ yea now I get it. It is very smart! (+1) Thanks for the explanation! $\endgroup$ – Shashi Feb 22 '17 at 21:36
  • $\begingroup$ @Shashi : You are welcome! :-) $\endgroup$ – user90369 Feb 23 '17 at 6:08

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.