Review question

# Can we find all three integrals? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5126

## Solution

1. Prove that $\begin{equation*} \int_0^{\frac{1}{2}\pi} \frac{\cos x}{3 + \cos^2 x} \:dx = \tfrac{1}{4} \ln 3. \end{equation*}$

If the denominator of the integrand were an expression involving $\sin x$ we might be able to integrate by substitution…

Since $\cos^2 x = 1 - \sin^2 x$, we can write $\begin{equation*} \int_0^{\frac{1}{2}\pi} \frac{\cos x}{3 + \cos^2 x} \:dx = \int_0^{\frac{1}{2}\pi} \frac{\cos x}{4 - \sin^2 x} \:dx. \end{equation*}$ Now using partial fractions we can expand, $\dfrac{1}{4-z^2} = \dfrac{1}{4}\left(\dfrac{1}{2-z}+\dfrac{1}{2+z}\right)$. Then we can integrate using the substitutions $u=2\pm\sin x$. \begin{align*} \int_0^{\frac{1}{2}\pi} \frac{\cos x}{4 - \sin^2 x} \:dx &= \dfrac{1}{4}\int_0^{\frac{1}{2}\pi} \frac{\cos x}{2 - \sin x}+\frac{\cos x}{2 + \sin x} \:dx\\ &= \dfrac{1}{4}\bigl[-\ln\big\vert 2-\sin x\big\vert +\ln \big\vert 2+\sin x\big\vert\bigr]_0^{\frac{1}{2}\pi}\\ &= \dfrac{1}{4}\left[\ln \bigg\vert \dfrac{2+\sin x}{2-\sin x}\bigg\vert\right]_0^{\frac{1}{2}\pi}\\ &= \dfrac{1}{4}\ln 3 \end{align*}
1. Integrate with respect to $x$ $\begin{equation*} (a)\; \frac{x}{(x-1)^3} \end{equation*}$

If the numerator were a constant, this would look much more straight-forward, so let’s try rearranging…

We can write, $\begin{equation*} \int \frac{x}{(x-1)^3} \:dx = \int \frac{x-1}{(x-1)^3} + \frac{1}{(x-1)^3} \:dx = \int \frac{1}{(x-1)^2} + \frac{1}{(x-1)^3} \:dx. \end{equation*}$ If $u = x-1$, then $\begin{equation*} \int \frac{1}{(x-1)^2} + \frac{1}{(x-1)^3} \:dx = \int u^{-2} + u^{-3} \:du = \frac{u^{-1}}{-1} + \frac{u^{-2}}{-2} + c = \frac{1}{1-x} - \frac{1}{2(x-1)^2} + c \end{equation*}$

where $c$ is a constant of integration.

1. Integrate with respect to $x$ $\begin{equation*} (b)\;(\ln x)^2. \end{equation*}$

The general formula for integration by parts is $\int u \frac{dv}{dx} \, dx = uv - \int v\frac{du}{dx} \, dx.$

If we let $u = (\ln x)^2$ and $v = x$ then $\frac{dv}{dx}=1$ and $\frac{du}{dx}=2\ln x\times\frac{1}{x}$. Integrating by parts, we see that

$$$\label{eq:ii-b-second-ibp} \int (\ln x)^2 \:dx = x (\ln x)^2 - 2 \int \ln x \:dx.$$$ In order to integrate $\ln x$, we again integrate by parts using $u = \ln x$ and $v = x$. Then $\frac{dv}{dx}=1$ and $\frac{du}{dx}=\frac{1}{x}$ and we have $\begin{equation*} \int \ln x \:dx = x \ln x - \int x \frac{1}{x} \:dx = x \ln x - x. \end{equation*}$ After substituting this into $\eqref{eq:ii-b-second-ibp}$, we can conclude that $\begin{equation*} \int (\ln x)^2 \:dx = x (\ln x)^2 - 2 \left( x \ln x - x \right) = x (\ln x)^2 - 2 x \ln x + 2x + c \end{equation*}$

where $c$ is, again, a constant of integration.