# Thread: trouble starting integration by parts

1. ## trouble starting integration by parts

$\displaystyle xtan^2(x)dx$ on [0, pi/3]. i don't know how to get this one going. the book says to use substitution before integration by parts, but i don't see what to substitute. i tried $\displaystyle 1+sec^2(x)$ for $\displaystyle tan^2(x)$ but i didn't get anywhere.

2. Originally Posted by vassago
$\displaystyle xtan^2(x)dx$ on [0, pi/3]. i don't know how to get this one going. the book says to use substitution before integration by parts, but i don't see what to substitute. i tried $\displaystyle 1+sec^2(x)$ for $\displaystyle tan^2(x)$ but i didn't get anywhere.
It should be $\displaystyle \sec^2 x-1=\tan^2 x$

The integral would become $\displaystyle \int_0^{\frac{\pi}{3}} \left(x\sec^2x-x\right)\,dx=\int_0^{\frac{\pi}{3}} x\sec^2x\,dx-\int_0^{\frac{\pi}{3}} x\,dx$

The first integral can then be done by parts, with $\displaystyle u=x$ and $\displaystyle \,dv=\sec^2x\,dx$

Can you take it from here?

3. Originally Posted by vassago
$\displaystyle xtan^2(x)dx$ on [0, pi/3]. i don't know how to get this one going. the book says to use substitution before integration by parts, but i don't see what to substitute. i tried $\displaystyle 1+sec^2(x)$ for $\displaystyle tan^2(x)$ but i didn't get anywhere.
Substitute $\displaystyle \tan^2 x = \sec^2 x - 1$. Then your integral becomes $\displaystyle \int x \sec^2 x \, dx - \int x \, dx$.

Now use integration by parts on the first integral: $\displaystyle u = x \Rightarrow du = dx$ and $\displaystyle dv = \sec^2 x \, dx \Rightarrow v = \tan x$.

Blue sky all the way now.