- Evaluate \[\begin{equation*} \int_0^{\tfrac{1}{4} \pi} \tan^3 \theta \:d\theta, \end{equation*}\] giving the answer correct to two significant figures.
So we have two integrals to calculate; the first is
\[\int_0^{\tfrac{1}{4} \pi} \sec^2 \theta\tan \theta \:d\theta\]
We notice that \(\dfrac{d}{d \theta} \tan \theta = \sec^2 \theta\) (this is a standard result). So, letting \(u = \tan \theta\), our integral becomes
\[\int_0^{\tfrac{1}{4} \pi} u \dfrac{du}{d\theta} \:d\theta = \int_0^{\tan \tfrac{\pi}{4}} u \:du = \dfrac{1}{2} \left[\dfrac{u^2}{2}\right]_0^{\tan \tfrac{\pi}{4}} = \dfrac{1}{2} - 0 = \dfrac{1}{2}\]
Our second integral is
\[\int_0^{\tfrac{1}{4} \pi} tan \theta \:d\theta = \int_0^{\tfrac{1}{4} \pi} \dfrac{\sin \theta}{\cos \theta} \:d\theta\]
We can calculate this by writing \(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\), substituting \(v = \cos \theta\), and noticing that \(-\sin \theta \:d\theta = \:dv\), so our integral is in fact
\[\int_0^{\tfrac{1}{4} \pi} \dfrac{-1}{v} \dfrac{\:d v}{\:d \theta} \:d\theta = \int_0^{\cos \tfrac{\pi}{4}} \dfrac{-1}{v} \:dv = - \biggl[ \ln v \biggr]_1^{\cos \tfrac{\pi}{4}} = \biggl[ \ln \dfrac{1}{v} \biggr]_1^{\tfrac{1}{\sqrt 2}} = \ln{\sqrt 2} - \ln 1 = \dfrac{1}{2} \ln 2 \]
So our final value for \(\int_0^{\tfrac{1}{4} \pi} \tan^3 \theta \:d\theta\) is
\[\dfrac{1}{2} - \dfrac{1}{2} \ln 2 = 0.1534...\]
which equals \(0.15\) to two significant figures.which equals \(0.15\) to two significant figures.
- By using the substitution \(x+1 = u^2\), or otherwise, find \[\begin{equation*} \int \frac{\sqrt{x+1}}{x-3} \:dx. \end{equation*}\]
Take \(x+1 = u^2\), so that \(x = u^2 - 1\), and \(dx = 2u du\).
Hence, \[\begin{align*} \int \frac{\sqrt{x+1}}{x-3} \:dx &= \int \frac{u}{u^2-4} 2u\:du \\ &= \int \frac{2u^2 - 8+8}{u^2 - 4} \:du\\ &= \int 2 + \frac{8}{(u-2)(u+2)} \:du. \end{align*}\] We can use partial fractions, \[\begin{align*} \frac{8}{(u-2)(u+2)} &= \frac{A}{u-2} + \frac{B}{u+2}\\ \implies\quad 8 &= A(u+2) + B(u-2) \end{align*}\] and substituting \(u=\pm 2\) we find \(A=2\) and \(B=-2\). Hence, \[\begin{align*} \int \frac{\sqrt{x+1}}{x-3} \:dx &= 2u + 2 \int \frac{1}{(u-2)} - \frac{1}{(u+2)} \:du \\ &= 2u + 2 \log(u-2) - 2\log(u+2) + c \\ &= 2\sqrt{x+1} + 2\log(\sqrt{x+1} - 2) - 2\log(\sqrt{x+1} + 2) + c. \end{align*}\]
