How do I solve INT x sec^2x dx using integration by parts?

[ryan750]

show that INT x sec^2x dx = pi/4 - ln2/2 (between pi/4 and 0)

pls help i don't know where to start

i know it is integration by parts - just don't know how i should rearrange it. thanks

 

[dextercioby]

If you know part integration,u'll find the notation quite familiar

$$x=u;\sec x dx =dv$$

BTW,it should read

$$\int_{0}^{\frac{\pi}{4}} x\sec^{2} x \ dx=\frac{\pi}{4}-\frac{\ln 2}{2}$$

Daniel.

 

[Jameson]

Tabular method, so much easier. Enough said.

 

[BobG]

Tabular method really only helps when you have several ibp steps. In this case, you only have one. Once you integrate sec^2 x, you get tan x. If you're going to integrate tan x manually (vs just looking at the table), it's easiest to break it into

$$\int{\sin x \frac{1}{\cos x} dx}$$

and then use u-substitution.