- By using the substitution \(u = 1 - x^2\), or otherwise, evaluate
\[\begin{equation*}
\int_0^{\frac{1}{2}} \frac{x}{(1-x^2)^2} \:dx.
\end{equation*}\]
If we take
\(u = 1 - x^2\), then
\[\begin{equation*}
\frac{du}{dx} = -2x \implies du = -2x \:dx
\end{equation*}\]
and therefore, by substitution,
\[\begin{align*}
\int_0^{\frac{1}{2}} \frac{x}{(1-x^2)^2} \:dx = \int_{1-0}^{1-\left(\frac{1}{2}\right)^2} \frac{du}{-2u^2} &= \frac{1}{2} \int_{\frac{3}{4}}^1 \frac{du}{u^2} \\
&= \frac{1}{2} \left[ \frac{u^{-1}}{-1} \right]_{\frac{3}{4}}^1 \\
&= -\frac{1}{2} \left( 1 - \frac{4}{3} \right) \\
&= \frac{1}{6}.
\end{align*}\]
Note how we can make an integral look simpler by factoring out constants, including negatives which we can take care of by swapping the limits over.
- Evaluate
\[\begin{equation*}
\int_0^{\pi/6} \frac{\sin \theta \:d\theta}{\cos^3 \theta}.
\end{equation*}\]
In part (i) the \(x^2\) in the substitution dealt with the \(x\) in the numerator. In the same way here we don’t need to worry about the \(\sin \theta\) if we have \(\cos \theta\) in the substitution—we can be confident it will get sorted out along the way.
Take
\(u = \cos \theta\), so that
\[\begin{equation*}
\frac{du}{d\theta} = -\sin \theta \implies du = - \sin \theta \:d\theta.
\end{equation*}\]
By substitution, then,
\[\begin{align*}
\int_0^{\pi/6} \frac{\sin \theta \:d\theta}{\cos^3 \theta} = \int_{\cos 0}^{\cos \frac{\pi}{6}} \frac{-du}{u^3} &= - \left[ \frac{u^{-2}}{-2} \right]_1^{\sqrt{3}/2} \\
&= \frac{1}{2} \left( \left( \frac{2}{\sqrt{3}} \right)^2 - 1 \right) \\
&= \frac{1}{6}.
\end{align*}\]
