1. Aug 8, 2012

### XtremePhysX

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

Evaluate the following integral:

2. Relevant equations

$$\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx$$

3. The attempt at a solution

I tried to split is into ln(1) times ln(tanx) but didnt work.

2. Aug 8, 2012

### Millennial

Because you can't split it up like that perhaps?

Use this identity of integrals: Substituting x=a-t, dx=-dt:

$$\int_{0}^{a}f(a-t)\,dt=-\int_{a}^{0}f(x)\,dx=\int_{0}^{a}f(x)\,dx$$

3. Aug 8, 2012

### XtremePhysX

can u please show me how to do it, i tried it this way but i messed it up.

4. Aug 8, 2012

### clamtrox

That is a pretty difficult integral. If you really wrote it down correctly, the best approach is likely to evaluate it numerically.

5. Aug 8, 2012

### Millennial

I'll give you a headstart, but you should solve it yourself.
Applying the result I gave above to your problem, we get that your integral is equal to
$$\int_{0}^{\pi/4}\log(1+\tan(\pi/4-x))dx$$
Now, note that $\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$ so that
$$\tan(\pi/4-x)=\frac{\sin(\pi/4-x)}{\cos(\pi/4-x)}=\frac{\sin(\pi/4)\cos(x)-\cos(\pi/4)\sin(x)}{\cos(\pi/4)\cos(x)+\sin(\pi/4)\sin(x)}=\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}$$
This is easily seen to be equal to
$$\frac{1-\tan(x)}{1+\tan(x)}$$
Now, please take it from there.

6. Aug 8, 2012

### Millennial

The indefinite integral is in terms of the logarithmic integral function but the definite integral has a solution.

7. Aug 8, 2012

### XtremePhysX

Thank u a lot.

great forum.

8. Aug 8, 2012

### Millennial

What does that mean? Do you have the answer??

9. Aug 8, 2012

### XtremePhysX

I don't have the answer but thanks to u, i solved it, the answer is pi/4 ln(2)

10. Aug 8, 2012

### XtremePhysX

guys, im learning harder integration so can any one give me an example of an extremely difficult integral?

Last edited: Aug 8, 2012
11. Aug 8, 2012

### Millennial

$$\int_{0}^{\infty}\frac{\cos(x)\,dx}{1+x^2}$$
Good luck!

12. Aug 8, 2012

### XtremePhysX

wow thats a hard one
do you have so really big, difficult and intimidating integrals that looks almost impossible to solve

13. Aug 8, 2012

### Millennial

Discussion here is not really appropriate, I shall send some to you via PM.

14. Aug 8, 2012

### XtremePhysX

thank you
im new here so excuse me please :)

15. Aug 9, 2012

### gabbagabbahey

Is it? You might want to sketch the integrand and see if your result is reasonable for the area under the curve ;0).

16. Aug 10, 2012

### srl17

Your really close, you probably made a simple arithmetic mistake in your work. The graph of that is very close to a triangle and the area you found wouldn't fit in the bounds of a triangle. That was fun integral though!

17. Aug 10, 2012

### Ray Vickson

Wrong. If we let $J = \int_0^{\pi/4} \, \ln(1 + \tan(x))\, dx,$ your claim is that $J = \pi/4\ln(2).$ If you mean $J = (\pi/4) \ln(2) \doteq 0.5443965,$ that is wrong. If you mean $J = \pi/(4 \ln(2)) \doteq 1.1330900,$ that is also wrong. The correct answer is
$$J = \frac{i}{2}\text{dilog}\left(\frac{1+i}{2}\right) - \frac{i}{2}\text{dilog}\left(\frac{1-i}{2}\right) + \frac{\pi \ln(2)}{4} - C,$$
where C is Catalan's constant
$$C = \sum_{i=0}^{\infty} \frac{(-1)^i}{(2i+1)^2}$$
and
$$\text{dilog}(w) = \int_1^w \frac{\ln(t)}{1-t} \, dt$$
and $i = \sqrt{-1}.$ The numerical value of the correct answer is
$J \doteq 0.27219826.$ (All this courtesy of Maple11).

RGV

18. Aug 10, 2012

### srl17

J≐0.27219826 looks like $$\frac{\pi}{8}\ln{2}$$ which is why I think it likely he made a simple arithmetic mistake.

19. Aug 10, 2012

### Bohrok

He probably made the same mistake I did when I worked it out and got the same answer as him but I did eventually get$$\frac{\pi}{8}\ln2$$