What is the best method for integrating x(arctanx)^2?

[Ayesh]

Homework Statement

I have to integrate using the following integration methods:
1) u-sub
2) by parts
3) trig
4) partial fraction

Homework Equations

\intx(arctanx)²

The Attempt at a Solution

According to me, the method I should use is integration by parts.

When I tried it, the whole thing got worse...

u = arctanx²
du = 2arctanx/1+x²
dv = xdx
v = 1/2x²

\intx(arctanx)² = 1/2x²(arctanx)² - \intx²arctanx/1+x₂

 

[Kreizhn]

I would use integration by parts, though you're going to need to use it twice.

First use IBP to find calculate \int \arctan(x) dx. Once you have this, you can use IBP on the whole thing.

 

[tt2348]

try using a U substitution, say u=arctan(x) and du=\frac{1}{1+x^{2}}dx
then dx=(1+x^{2})du , where x=tan(u). also plugging in for the x multiplied in front gets you \int u^{2}tan(u)(1+tan(u)^{2})du

 

[Kreizhn]

try using a U substitution, say u=arctan(x) and du=\frac{1}{1+x^{2}}dx
then dx=(1+x^{2})du , where x=tan(u). also plugging in for the x multiplied in front gets you \int u^{2}tan(u)(1+tan(u)^{2})du

That is definitely more elegant, though I'm not sure that Ayesh will see the trick involved to make it an easy IBP.

 

[Ayesh]

Kreizhn why should I use \intarctanx first?

 

[Ayesh]

I get from where dx(1 + tan(u)²)du comes from, but not the u²tan(u).

 

[Kreizhn]

If you're going to keep going in the more elegant direction, I wouldn't worry about integrating arctan first. It is a method that works, but takes a lot more work.

As for your other question, consider that your integrand is x(\arctan(x))^2 and you make the substitution u = \arctan(x) or alternatively, x = \tan(u). Now when making the substitution, there are three things you are going to need to account for, namely, the x, the \arctan^2(x) and the dx.

see if you can follow the following bit of arithmetic

\begin{align*}<br /> x \arctan^2(x) dx &amp;= x u^2 dx &amp; \text{ since } u=\arctan(x) \\<br /> &amp;= tan(u) u^2 dx &amp; \text{ since } x = \tan(u) \\<br /> &amp;= u^2 \tan(u) (1+x^2) du &amp; \text{ since } dx = (1+x^2) du \\<br /> &amp;= u^2 \tan(u) (1+tan^2(u) )du &amp; \text{ since } x = \tan(u)<br /> \end{align*}<br />

The tricky part now is realizing that there is an implicit derivative in here that will make integration by parts simple.

 

[Ayesh]

Thank you!

Your explanations are very clear!

 

[Ayesh]

This is what I have done until now:

\intx(arctanx)^2 dx

1/2 x²arctanx - integral 1/2x²(1/1+x^2) dx

1/2x²arctanx - 1/2 integral tan²t/1+tan²t *sec²t

1/2x²arctanx - 1/2 integral tan²t

1/2x²arctanx - 1/2 integral (sec²t - 1) dt

1/2x²arctanx - 1/2 integral (tant - t) dt

... ?

I don't what to do after.
I know it has something to do with drawing a triangle, but I don't know what values to put around it.