# How do you integrate int xsec^2x by parts from [0, pi/4]?

May 25, 2017

$\frac{1}{4} \pi - \frac{1}{2} \ln 2$

#### Explanation:

First without the bounds:

$\int x {\sec}^{2} x \textcolor{w h i t e}{.} \mathrm{dx}$

Since integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$, we need to assign values for $u$ and $\mathrm{dv}$. Let:

$\left\{\begin{matrix}u = x & \implies & \mathrm{du} = \mathrm{dx} \\ \mathrm{dv} = {\sec}^{2} x \textcolor{w h i t e}{.} \mathrm{dx} & \implies & v = \tan x\end{matrix}\right.$

So:

$\int x {\sec}^{2} x \textcolor{w h i t e}{.} \mathrm{dx} = x \tan x - \int \tan x \textcolor{w h i t e}{.} \mathrm{dx}$

You may know this integral already, but if you're unsure how to find it, the process goes as such:

$- \int \tan x \textcolor{w h i t e}{.} \mathrm{dx} = \int \frac{- \sin x}{\cos} x \mathrm{dx}$

Let $t = \cos x$ so $\mathrm{dt} = - \sin x \textcolor{w h i t e}{.} \mathrm{dx}$:

$= \int \frac{\mathrm{dt}}{t} = \ln \left\mid t \right\mid = \ln \left\mid \cos \right\mid x$

So:

$\int x {\sec}^{2} x \textcolor{w h i t e}{.} \mathrm{dx} = x \tan x + \ln \left\mid \cos \right\mid x$

Applying the bounds:

${\int}_{0}^{\frac{\pi}{4}} x {\sec}^{2} x \textcolor{w h i t e}{.} \mathrm{dx} = {\left[x \tan x + \ln \left\mid \cos \right\mid x\right]}_{0}^{\frac{\pi}{4}}$

$= \frac{\pi}{4} \tan \left(\frac{\pi}{4}\right) + \ln \left\mid \cos \right\mid \left(\frac{\pi}{4}\right) - \left(0 \tan 0 + \ln \left\mid \cos \right\mid 0\right)$

$= \frac{\pi}{4} \left(1\right) + \ln \left\mid \frac{1}{\sqrt{2}} \right\mid + \left(0 + \ln \left\mid 1 \right\mid\right)$

Using $\frac{1}{\sqrt{2}} = {2}^{- \frac{1}{2}}$ and $\ln \left({a}^{b}\right) = b \ln \left(a\right)$:

$= \frac{\pi}{4} - \frac{1}{2} \ln 2$