Evaluating $\int_0^1\frac{\arctan t}{t\sqrt{1-t^2}}dt$

[dirk_mec1]

Homework Statement

Evaluate

$$\int_0^1\frac{\arctan t}{t\sqrt{1-t^2}}dt$$

The Attempt at a Solution

I suspect I need to subsitute something but what?

 

[CompuChip]

I'm getting a little lost in trig identities, so I haven't looked into it very much. But have you tried t = tan(x) to get rid of the arctan, or t = sin(x) or cos(x) to get rid of the square root ?

 

[dirk_mec1]

I'm getting a little lost in trig identities, so I haven't looked into it very much. But have you tried t = tan(x) to get rid of the arctan, or t = sin(x) or cos(x) to get rid of the square root ?

I've tried your suggestions and failed to get an integral which I can solve...

 

[n!kofeyn]

I think I have a solution. I may be doing it by a method more complicated than required, but sometimes these integrals refuse to be done easily. I will give hints, and I can give more details if you need them.

With integrals that contain this type of square root on the bottom, you will almost always do a trigonometric substitution. The two identities that are useful are [itex]cos^2\theta + \sin^2\theta = 1[/tex] and $1+\tan^2\theta=\sec^2\theta$. Since we have $$1-t^2$$[/itex][tex], the first identity is what we want to use. Let $t=\sin\theta$ so that you won't get negative signs when differentiating. Do this substitution and then simplify. Go ahead and post the integral that you get after this, and then we'll move on.[/tex]

 

[CompuChip]

Hmm the answer comes out so nicely one thinks there must be some good substitution.
But I'm seeing your problem: you are just two steps away from the solution but whatever you try seems to bring you just one step closer but no more.

Actually I tried calculating the integral with Mathematica, the funny thing is that it gives an exact answer precisely for boundaries 0 and 1 and not, for example for 1/2 to 1 or 0 to 2. So probably there is a sneaky argument in there somewhere, although I am unable to see where the answer is coming from, except possible from the substitution t = sinh(x) but I don't see how that would help.

In any case, I don't really know how to help you at this point, I've run out of possible ways to solve it. I hope someone else might be able to help you.

[edit]n!kofeyn is making me curious, I'm following closely :) [/edit]

 

[snipez90]

Hmmm I also pursued n!kofeyn's first step but I'm lost on how to deal with arctan(sinx). I can work with the limits of integration but there must be another clever step (or maybe easy) that I'm missing. I'll look at this later.

 

[FedEx]

Finally i scored one over a Homework Helper.

Phew! That was good.

Take $$t= tan\theta$$

Bring the entire integral in terms of sines and cosines.

Then substitute $$cos2\theta = {\alpha}^2$$

And then solve it by partial fractions.

PS Resolving to partial is quite tricky. And even after that there are more substitutions to come

 

[dirk_mec1]

Ok I got (leaving out constants and integration limits):$$\int \frac{\theta} { \sin( \theta) \sqrt{ \cos( 2 \theta)}}\ \mbox{d} \theta$$

 

[FedEx]

Its cos(2Theta) and not cos(theta)

Oh crap! I skipped a step.

$$cos2\theta = {\alpha}^2$$

You will have to apply parts before the above substitution. So that we will get rid of the $$\theta$$

 

[dirk_mec1]

Its cos(2Theta) and not cos(theta)

Yes I've editted it. Now your substitution alpha^2 =cos (2* theta) is nasty are you sure this is right?

 

[FedEx]

After doing that bring the entire term in terms of alpha. You will get

$$\int\frac{1}{(\frac{1-{\alpha}^2}{\alpha})(\sqrt{{\alpha}^2 +1})}$$

Do this with partial fractions.

PS I have not written the 2's and root 2's coming in the equation

 

[dirk_mec1]

After doing that bring the entire term in terms of alpha. You will get

$$\int\frac{1}{(\frac{1-{\alpha}^2}{\alpha})(\sqrt{{\alpha}^2 +1})}$$

Do this with partial fractions.

PS I have not written the 2's and root 2's coming in the equation

The theta = 0.5*cos^-1(alpha^2) where's that term?

 

[CompuChip]

You will have to apply parts before the above substitution. So that we will get rid of the $$\theta$$

You get rid of it in the integration by parts, I presume.

 

[dirk_mec1]

In the IBP: integrate w.r.t. to what?

 

[Cyosis]

I am pretty sure FedEx has made a mistake integrating by parts and as a result his method seems to lead to a dead end. CompuChip's analysis with mathematica should give us a clue that there exists no primitive to the integral. However there is a smart trick to get an exact value for the definite integral from 0 to 1.

First we define the function $$f(a) = \int_0^1 \frac{\arctan(a t)}{t \sqrt{1-t^2}}$$ with $$a \geq 0$$. Then take the derivate of f(a) with respect to a and use the substitution $$t=\frac{1}{\sqrt{1+u^2}}$$.

$$\begin{align*}<br /> f'(a) & = \int_0^1 \frac{1}{a^2 t^2+1} \frac{1}{\sqrt{1-t^2}}\,dt<br /> \\<br /> & = - \int_{\infty}^0 \frac{1}{(\frac{a^2}{1+u^2}+1) \sqrt{1 - \frac{1}{1+u^2}}} \frac{u}{(1+u^2)^{\frac{3}{2}}}\,du<br /> \\<br /> &= \int_0^{\infty} \frac{du}{a^2+1+u^2}<br /> <br /> \end{align*}$$

You should be able to rearrange the integrand so you recognize the arctangent. Note that we now have f'(a). Now try to find the answer of your original integral with a=1.

 

[FedEx]

You get rid of it in the integration by parts, I presume.

Precisely

 

[FedEx]

In the IBP: integrate w.r.t. to what?

With respect to alpha.

 

[Cyosis]

You will have to apply parts before the above substitution. So that we will get rid of the $$\theta$$

FedEx could you show us how you did the integration by parts. I have tried to get rid of $$\theta$$ as you suggest, but it doesn't seem to lead to a manageable integral.

 

[dirk_mec1]

FedEx could you show us how you did the integration by parts. I have tried to get rid of $$\theta$$ as you suggest, but it doesn't seem to lead to a manageable integral.

I agree, Fedex you've made mistake somewhere.

 

[dirk_mec1]

//

You should be able to rearrange the integrand so you recognize the arctangent. Note that we now have f'(a). Now try to find the answer of your original integral with a=1.

Nice! I'll look carefully at your derivation.

 

[FedEx]

I agree, Fedex you've made mistake somewhere.

Yes i had.A very grave one. But finally got the answer. My method is unncecessary lenghty. Complicated. But here it goes

http://s557.photobucket.com/albums/ss20/FedExMilind/

I request all of you never use the method which i have. Its very long. The only consolation which i can fetch is that i got the answer. I am still not sure whether the solution is correct or not. But just have look at it.

I had completely fogotten a term to write. Which led to the blunder.

 

[FedEx]

I am pretty sure FedEx has made a mistake integrating by parts and as a result his method seems to lead to a dead end. CompuChip's analysis with mathematica should give us a clue that there exists no primitive to the integral. However there is a smart trick to get an exact value for the definite integral from 0 to 1.

First we define the function $$f(a) = \int_0^1 \frac{\arctan(a t)}{t \sqrt{1-t^2}}$$ with $$a \geq 0$$. Then take the derivate of f(a) with respect to a and use the substitution $$t=\frac{1}{\sqrt{1+u^2}}$$.

$$\begin{align*}<br /> f'(a) & = \int_0^1 \frac{1}{a^2 t^2+1} \frac{1}{\sqrt{1-t^2}}\,dt<br /> \\<br /> & = - \int_{\infty}^0 \frac{1}{(\frac{a^2}{1+u^2}+1) \sqrt{1 - \frac{1}{1+u^2}}} \frac{u}{(1+u^2)^{\frac{3}{2}}}\,du<br /> \\<br /> &= \int_0^{\infty} \frac{du}{a^2+1+u^2}<br /> <br /> \end{align*}$$

You should be able to rearrange the integrand so you recognize the arctangent. Note that we now have f'(a). Now try to find the answer of your original integral with a=1.

This is pure Genious

 

[n!kofeyn]

Hmmm I also pursued n!kofeyn's first step but I'm lost on how to deal with arctan(sinx). I can work with the limits of integration but there must be another clever step (or maybe easy) that I'm missing. I'll look at this later.

Sorry snipez90 and CompuChip for the delay, but I had made a mistake in my next step. I tried using integration by parts to get rid of the $\arctan(\sin\theta)$, but this doesn't work. Another technique I tried is to repeatedly apply integration by parts to the original integral, trying to arrive back at the same integral. Then you could use the trick of having Integral = some stuff - Integral, so that you then have 2Integral = some stuff, but unless I missed a negative sign, this did not work.