- By using the substitution \(x = \cos \theta\), or otherwise, evaluate
\[\begin{equation*}
\int_0^1 \frac{1}{(1-x)^{1/2}(1+x)^{3/2}} \:dx.
\end{equation*}\]
If
\(x = \cos \theta\), then
\(dx = -\sin \theta \:d\theta\), and so
\[\begin{align*}
\int_0^1 \frac{1}{(1-x)^{1/2}(1+x)^{3/2}} \:dx &= \int_{\pi/2}^0 \frac{-\sin \theta}{(1 - \cos \theta)^{1/2}(1 + \cos \theta)^{3/2}} \:d\theta \\
&= \int_0^{\pi/2} \frac{\sin \theta}{(1 - \cos \theta)^{1/2}(1 + \cos \theta)^{3/2}} \:d\theta \\
&= \int_0^{\pi/2} \frac{\sin \theta}{(1 - \cos \theta)^{1/2}(1 + \cos \theta)^{1/2}(1 + \cos \theta)} \:d\theta \\
&= \int_0^{\pi/2} \frac{\sin \theta}{(1 - \cos^2 \theta)^{1/2}(1 + \cos \theta)} \:d\theta \\
&= \int_0^{\pi/2} \frac{d\theta}{1 + \cos \theta}.
\end{align*}\]
From the
double angle formula for cosine, we have that
\[\begin{equation*}
\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = 2 \cos^2 \frac{\theta}{2} - 1 \implies 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}.
\end{equation*}\]
Hence,
\[\begin{align*}
\int_0^{\pi/2} \frac{d\theta}{1 + \cos \theta} &= \frac{1}{2} \int_0^{\pi/2} \sec^2 \frac{\theta}{2} \:d\theta\\
&=\left[\tan \dfrac{\theta}{2}\right]_0^{\pi/2}\\
&= \tan \frac{\pi}{4} - \tan 0 = 1.
\end{align*}\]
- Find
\[\begin{align*}
\text{(i)}\; &\int \frac{1}{\sqrt{x+2} - \sqrt{x+1}} \:dx, \\
\end{align*}\]
Rationalising the denominator, we have that
\[\begin{align*}
\int \frac{1}{\sqrt{x+2} - \sqrt{x+1}} \:dx &= \int \frac{\sqrt{x+2} + \sqrt{x+1}}{(\sqrt{x+2} - \sqrt{x+1})(\sqrt{x+2} + \sqrt{x+1})} \:dx \\
&= \int \frac{\sqrt{x+2} + \sqrt{x+1}}{(x+2) - (x+1)} \:dx \\
&= \int \sqrt{x+2} + \sqrt{x+1} \:dx \\
&= \frac{(x+2)^{3/2}}{\frac{3}{2}} + \frac{(x+1)^{3/2}}{\frac{3}{2}} + c \\
&= \frac{2}{3} \left( (x+2)^{3/2} + (x+1)^{3/2} \right) + c
\end{align*}\]
where \(c\) is a constant of integration.
\[\begin{align*}
\text{(ii)}\; &\int \frac{x}{(x+1)(x+2)^2} \:dx.
\end{align*}\]
We can attempt to rewrite the integrand using partial fractions: that is, we would hope that there are constants
\(A\),
\(B\), and
\(C\) such that
\[\begin{equation*}
\frac{x}{(x+1)(x+2)^2} = \frac{A}{x+1} + \frac{B}{x+2} +\frac{C}{(x+2)^2}= \frac{A(x+2)^2 + B(x+1)(x+2)+C(x+1)}{(x+1)(x+2)^2}.
\end{equation*}\]
Thus \(x = A(x+2)^2 + B(x+1)(x+2)+C(x+1)\) for all values of \(x\).
Putting \(x = -1\), we have \(A = -1\), while putting \(x = -2\) gives \(C=2\) and putting \(x = 0\) yields \(B = 1\).
As a result,
\[\begin{align*}
\frac{x}{(x+1)(x+2)^2} &= \frac{1}{x+2} + \frac{2}{(x+2)^2} - \frac{1}{x+1}
\end{align*}\]
and so
\[\begin{align*}
\int \frac{x}{(x+1)(x+2)^2} \:dx &= \int \frac{1}{x+2} + \frac{2}{(x+2)^2} - \frac{1}{x+1} \:dx \\
&= \ln\big\vert x+2\big\vert - \frac{2}{x+2} - \ln\big\vert x+1\big\vert + c
\end{align*}\]
where \(c\) is a constant of integration.
