integration question, please help!

XtremePhysX

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

Evaluate the following integral:

2. Relevant equations

[tex]\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx[/tex]

3. The attempt at a solution

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

XtremePhysX

bump

Millennial

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

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

[tex]\int_{0}^{a}f(a-t)\,dt=-\int_{a}^{0}f(x)\,dx=\int_{0}^{a}f(x)\,dx[/tex]

XtremePhysX

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

clamtrox

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

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
[tex]\int_{0}^{\pi/4}\log(1+\tan(\pi/4-x))dx[/tex]
Now, note that [itex]\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}[/itex] so that
[tex]\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)}[/tex]
This is easily seen to be equal to
[tex]\frac{1-\tan(x)}{1+\tan(x)}[/tex]
Now, please take it from there.

Millennial

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

XtremePhysX

Thank u a lot.

great forum.

Millennial

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

XtremePhysX

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

XtremePhysX

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

Millennial

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

XtremePhysX

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

Millennial

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

XtremePhysX

thank you
im new here so excuse me please :)

gabbagabbahey

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

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!