- 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…
- 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…
where \(c\) is a constant of integration.
- 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.
