How do you integrate #tanxtan2x#?

I did it a bit differently, and I also got #-x + 1/2ln|(1 + tanx)/(1 - tanx)| + C#.

If you're like me, and you don't remember the trig identity for #tan(2x)# (assuming you mean that), you can simply rewrite this in terms of #sin# and #cos#...

#int (sinxsin(2x))/(cosxcos(2x))dx#

#= int (sinx (2sinxcancel(cosx)))/(cancel(cosx)(cos^2x - sin^2x))dx#

#= 2 int (sin^2x)/(cos^2x - sin^2x)dx#

#= 2 int (sin^2x - cos^2x + cos^2x)/(cos^2x - sin^2x)dx#

#= 2 int -1 dx + 2 int cos^2x/(cos^2x - sin^2x)dx#

#= -2x + 2 int cos^2x/(cos^2x - sin^2x)dx#

#= -2x + 2 int 1/(1 - tan^2x)dx#

Now, let #u = tanx# so that #du = sec^2xdx#. Then, #dx = 1/(sec^2x)du = 1/(1 + u^2)du# and:

#= -2x + 2 int 1/((1 - u^2)(1 + u^2))du#

Partial fraction decomposition gives:

#int 1/((1 - u^2)(1 + u^2))du = int A/(1 + u) + B/(1 - u) + (Cu + D)/(1 + u^2)du#

You should get:

• #A = -B = -1/4#,
• #C = 0#,
• #D = 1/2#,

which means:

#2int 1/((1 - u^2)(1 + u^2))du = 2[ -1/4ln|1 - u| + 1/4 ln|1 + u| + 1/2 arctan(u)]#

And our result is:

#color(blue)(int tanxtan(2x)dx) = -2x + 2[1/4ln|(1 + tanx)/(1 - tanx)| + x/2] + C#

#= color(blue)(-x + 1/2ln|(1 + tanx)/(1 - tanx)| + C)#