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

\[\begin{equation}\label{eq:ii-b-second-ibp} \int (\ln x)^2 \:dx = x (\ln x)^2 - 2 \int \ln x \:dx. \end{equation}\] 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.