I seen this problem a while back on the site and to my recollection it wasn't tackled. If it was, my apologies. I couldn't find it. I think Isomorphism had posted it. If I am a for doing this, so be it.

I am sure some of you have some clever show off methods, but here is something that kind of appeared to me rather quickly and turned into something familiar. I think it is cool to see the different innovative ways folks tackle these. It shows what a wonder the human brain can be.

You can easily run them through a computer, but they lack the human touch.

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

Let $\displaystyle u=1+tan(x)$

$\displaystyle u-1=tan(x)dx$

$\displaystyle du=sec^{2}(x)dx$

Make the subs and get:

$\displaystyle \int\frac{ln(u)}{sec^{2}(x)}dx$

An x mixed in with a u?. No.......to some that may seem like sacrilege.

But it's OK:

Since $\displaystyle tan(x)=\frac{u-1}{1}$

Then $\displaystyle sec^{2}(u)=(u-1)^{2}+1$

Then we have:

$\displaystyle \int_{1}^{2}\frac{ln(u)}{(u-1)^{2}+1}du$

Now, let $\displaystyle w=u-1, \;\ w+1=u, \;\ dw=du$

Then we have:

$\displaystyle \int_{0}^{1}\frac{ln(w+1)}{w^{2}+1}dw$

Now...this is a familiar integral that the Krizmeister done awhile back so I won't bother.

It's solution is $\displaystyle \frac{\pi}{8}ln(2)$

This kind of popped into my head when I seen it so I had to write it down.