- Find
- \(\displaystyle \int x \sqrt{2x-1} \:dx\),
Let
\(u = 2x - 1\), so that
\(du = 2 \: dx\). Then
\[\begin{align*}
\int x \sqrt{2x-1} \:dx = \int \frac{u+1}{2} \sqrt{u} \:\frac{du}{2} &= \frac{1}{4} \int u^{3/2} + u^{1/2} \:du \\
&= \frac{1}{4} \left( \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{3/2}}{\frac{3}{2}} + C \right) \\
&= \frac{1}{10} u^{5/2} + \frac{1}{6} u^{3/2} + C\\
&= \frac{1}{10} (2x-1)^{5/2} + \frac{1}{6} (2x-1)^{3/2} + C,\\
\end{align*}\]
where \(C\) is a constant of integration.
- \(\displaystyle \int \frac{1}{\sqrt{2x - x^2}} \:dx\).
We can write
\[\begin{equation*}
2x - x^2 = -(x^2 - 2x) = -\left( (x-1)^2 - 1 \right) = 1 - (x-1)^2
\end{equation*}\]
so that
\[\begin{equation*}
\int \frac{1}{\sqrt{2x - x^2}} \:dx = \int \frac{1}{\sqrt{1 - (x-1)^2}} \:dx.
\end{equation*}\]
Take
\(\sin \theta = x - 1\), so that
\(\cos \theta \: d\theta = dx\) and
\[\begin{align*}
\int \frac{1}{\sqrt{1 - (x-1)^2}} \:dx = \int \frac{\cos \theta}{\sqrt{1 - \sin^2 \theta}} \:d\theta &= \int \frac{\cos \theta}{\cos \theta} \:d\theta \\
&= \theta + C \\
&= \sin^{-1} (x-1) + C,
\end{align*}\]
where \(C\) is a constant of integration.
- Evaluate \(\displaystyle \int_0^{\pi/4} \sec^4 x \:dx\).
We have that
\[\begin{equation*}
\tan^2 x + 1 = \sec^2 x.
\end{equation*}\]
By the standard result (which can be proved using the
quotient rule), we have that
\[\begin{equation*}
\frac{d}{dx} \tan x = \sec^2 x.
\end{equation*}\]
and therefore
\[\begin{equation*}
\int \sec^4 x \:dx = \int \sec^2 x \sec^2 x\:dx = \int \sec^2 x\tan^2 x + \sec^2 x \:dx.
\end{equation*}\]
The second term we can integrate by the standard result above. For the first term, we substitute \(u=\tan x\) and get \[
\int \sec^2 x\tan^2 x\:dx = \int u^2\:du = \frac{\tan^3 x}{3}+C.
\] Thus we have \[\int_0^{\pi/4} \sec^4 x \:dx = \left[\frac{\tan^3 x}{3}+\tan x\right]_0^{\pi/4} = \frac{4}{3}.
\]
