# Evaluate the following integral:

1. Dec 5, 2012

### Synchronised

1. The problem statement, all variables and given/known data

Evaluate the following integral using the substitution $$u=\frac{1-x}{1+x}$$

2. Relevant equations

$$\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx$$

3. The attempt at a solution

I tried using the given substitution but couldn't do the integral but I did it using $$u=tan^{-1}x$$ and got the answer = $$\frac{\pi ln2}{8}$$ so how can I do it with the given substitution?

Last edited: Dec 5, 2012
2. Dec 5, 2012

### SammyS

Staff Emeritus
Using your substitution, u=tan-1(x), what did you do with the ln(1+x) ?

I suspect that there is a typo in that integral . The suggested substitution, $\displaystyle \ \ u=\frac{1-x}{1+x},\ \$ works quite well to evaluate the integral, $\displaystyle \ \ \int_{0}^{1}\frac{\ln(1+x)}{(1+x)^{2}}\,dx\ .$

3. Dec 5, 2012

### Synchronised

No, I am sure there is no typo in the integral.
The way I did it was: $$Let u=tan^{-1}x\therefore dx=(1+x^{2})du\therefore\int_{0}^{\frac{\pi }{4}}ln(1+tanu)du=\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx=\int_{0}^{\frac{\pi }{4}}ln(1+tan(\frac{\pi }{4}-x))dx=$$
$$\int_{0}^{\frac{\pi }{4}}ln(1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx})dx=\int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-tanx}{1+tanx})dx=$$
$$\int_{0}^{\frac{\pi }{4}}ln(\frac{2}{1+tanx})dx=\int_{0}^{\frac{\pi }{4}}ln(2)dx-\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx$$
$$\therefore 2\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx=\int_{0}^{\frac{\pi }{4}}ln(2)dx$$
$$\therefore \int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx=\frac{\pi ln2}{8}$$

Last edited: Dec 5, 2012
4. Dec 5, 2012

### Millennial

The integration shown above is correct. I also verified it using Mathematica just in case.

The given substitution does not seem very well to me. Your substitution, to be honest, would be my first try.

5. Dec 5, 2012

### Synchronised

But the question says use the given substitution so I don't think I can get full marks if any by using a different substitution in the exam. Maybe the given substitution has something to do with $$\int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-tanx}{1+tanx})dx$$ ? they look similar.

6. Dec 5, 2012

### Synchronised

That looks better :)

Last edited: Dec 5, 2012
7. Dec 5, 2012

### Dick

To use the given substitution, just solve u=(1-x)/(1+x) for x. Then substitute that into ln(1+x). Yes, it does look pretty similar.

8. Dec 5, 2012

### Synchronised

Ok, thank you :)

$$Letu=tan^{-1}x\therefore dx=(1+x^{2})du\therefore\int_{0}^{\frac{\pi }{4}}ln(1+x)du$$
$$Let u=\frac{1-x}{1+x}\therefore x=\frac{1-u}{1+u}\therefore \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du...$$

Last edited: Dec 5, 2012
9. Dec 5, 2012

### SammyS

Staff Emeritus
The suggested substitution works perfectly well, when used right at the outset.

Your method has a curious step ...

$\displaystyle \int_{0}^{\pi /4}\ln(1+tanx)dx=\int_{0}^{\pi /4}\ln(1+tan(\frac{\pi }{4}-x))dx$

Of course, $\displaystyle \ \tan(x)\ne\tan(\pi/4-x)\,,\$ so what gives? This step can be shown to be valid by using the substitution, t = π/4 - x, then changing to variable of integration back to x.

10. Dec 5, 2012

### Dick

It looks like the sort of trick you might pull to evaluate a definite integral when there is no easy antiderivative. Integrate backwards and apply a trig identity. It's just being clever. Too clever. The given substitution gives you an indefinite integral without the need to resort to tricks like that.

11. Dec 6, 2012

### Synchronised

It's just a method to evaluate definite integrals. $$\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$$ to prove it you just need to use a dummy variable let u=a-x. therefore x=a-u and du=-dx $$\therefore \int_{0}^{a}f(x)dx=\int_{a}^{0}f(a-u)(-du)=\int_{0}^{a}f(a-u)(du)=\int_{0}^{a}f(a-x)(dx)$$

Last edited: Dec 6, 2012
12. Dec 6, 2012

### haruspex

I don't get that. I get a factor 1/u in the integrand.
dx = -2(1+u)-2du; (1+x2)-1 = (1+u)2/(4u); (1+x2)-1dx = -du/(2u).

13. Dec 6, 2012

### Synchronised

yeah you're right, i still can't evaluate the integral using the given sub from the start...

14. Dec 7, 2012

### SammyS

Staff Emeritus
I overlooked the above post earlier.

I hope you didn't mean that your original integral was equivalent to $\displaystyle \ \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du\ .$

Those are two different substitutions.

Using the suggested substitution:

$\displaystyle x=\frac{1-u}{1+u}=\frac{2}{1+u}-1\ \$ so that $\displaystyle \ \ x+1=\frac{2}{1+u}\ .$

Also, $\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=\frac{1+u^2}{(1+u)^2}\ .$

Put that together with what haruspex gave you for dx & see where that all leads you.

There is a mistake in my last line. It should be:

$\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .$

Last edited: Dec 7, 2012
15. Dec 7, 2012

### Synchronised

I get $$2\int_{0}^{1}\frac{ln(\frac{2}{1+u})}{1+u^{2}}du$$
So where do I go from here?

Last edited: Dec 7, 2012
16. Dec 7, 2012

### haruspex

You can write ln(2/(1+u)) a little differently. Then you'll see that one part of the integral looks very like what you started with, but with a constant multiplier.

17. Dec 7, 2012

### Synchronised

I almost got it, I get $$\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx=\frac{2ln2}{3}\int_{0}^{1}\frac{dx}{1+x^{2}} =\frac{\pi ln2}{6}$$ so I made a mistake somewhere :(

18. Dec 7, 2012

### haruspex

I think you got a sign wrong, leading to the divisor of 3.

19. Dec 7, 2012

### Synchronised

what should the divisor be? I keep getting 3.

Edit: It should be 4 but I really can't see the mistake, I checked the signs and everything...

Last edited: Dec 7, 2012
20. Dec 7, 2012

### SammyS

Staff Emeritus
I edited post #14, so divide the above by 2.

That should give you the same answer you originally got.