The Weierstrass Substitution (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam questions don't say to "use a Weierstrass substitution" but they do give the $t$-substitution in the question.

Use the substitution $t=\tan \left(\frac{\theta }{2}\right)$ to show that

$\int \sec \theta \mathrm{d}\theta =\ln \left|\begin{array}{rcl}& & \frac{1+\tan \left(\frac{\theta }{2}\right)}{1-\tan \left(\frac{\theta }{2}\right)}\end{array}\right|+c$

Answer:

Write $\sec \theta$ in terms of the cosine t-formula, $\cos \theta =\frac{1-t^2}{1+t^2}$

$\begin{array}{rcl}\sec \theta & =& \frac{1}{\cos \theta }\\ \sec \theta & =& \frac{1+t^2}{1-t^2}\end{array}$

Change $d\theta$ into $\mathrm{d}t$ by differentiating $t=\tan \left(\frac{\theta }{2}\right)$

$\frac{\mathrm{d}t}{\mathrm{d}\theta }=\frac{1}{2}{\sec }^2\left(\frac{\theta }{2}\right)$

Use $1+{\tan }^{2}A\equiv {\sec }^{2}A$ to write it in terms of $t$

$\begin{array}{rcl}\frac{\mathrm{d}t}{\mathrm{d}\theta }& =& \frac{1}{2}\left(1+{\tan }^2\left(\frac{\theta }{2}\right)\right)\\ \frac{\mathrm{d}t}{\mathrm{d}\theta }& =& \frac{1}{2}\left(1+t^2\right)\end{array}$

Find $d\theta$ in terms of $\mathrm{d}t$

$\begin{array}{rcl}\frac{\mathrm{d}\theta }{\mathrm{d}t}& =& \frac{2}{1+t^2}\\ \mathrm{d}\theta & =& \frac{2}{1+t^2}\mathrm{d}t\end{array}$

Substitute $\sec \theta$ and $d\theta$ into the integral

$\begin{array}{rcl}\int \sec \theta \mathrm{d}\theta & =& \int \frac{\cancel{1+t^2}}{1-t^2}\times \frac{2}{\cancel{1+t^2}}\mathrm{d}t\\ & =& \int \frac{2}{1-t^2}\mathrm{d}t\end{array}$

Write $\frac{2}{1-t^2}$ in partial fractions

$\begin{array}{rcl}\frac{2}{1-t^2}& =& \frac{A}{1+t}+\frac{B}{1-t}\\ 2& =& A\left(1-t\right)+B\left(1+t\right)\end{array}$

Find $A$ and $B$

$\begin{array}{rcl}t& =& 1\Rightarrow 2=2B\Rightarrow B=1\\ t& =& -1\Rightarrow 2=2A\Rightarrow A=1\end{array}$

Substitute the partial fractions into the integral

$\begin{array}{rcl}\int \sec \theta \mathrm{d}\theta & =& \int \frac{2}{1-t^2}\mathrm{d}t\\ & =& \int \left(\frac{1}{1+t}+\frac{1}{1-t}\right)\mathrm{d}t\end{array}$

Integrate each partial fraction

$\begin{array}{rcl}\int \sec \theta \mathrm{d}\theta & =& \ln |1+t|-\ln |1-t|+c\end{array}$

Use log laws to combine the two log terms

$\begin{array}{rcl}\int \sec \theta \mathrm{d}\theta & =& \ln \left|\frac{1+t}{1-t}\right|+c\end{array}$

Write the answer back in terms of $\theta$ using $t=\tan \left(\frac{\theta }{2}\right)$

$\begin{array}{rcl}\int \sec \theta \mathrm{d}\theta & =& \ln \left|\frac{1+\tan \left(\frac{\theta }{2}\right)}{1-\tan \left(\frac{\theta }{2}\right)}\right|+c\end{array}$

Use the substitution $t=\tan \left(\frac{\theta }{2}\right)$ to determine the exact value of

${\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2+2\cos \theta +\sin \theta }d\theta$

giving your answer in the form $\ln \left(p+q\sqrt{3}\right)$, where $p$ and $q$ are constants to be found.

Answer:

Use the t-formulae $\cos \theta =\frac{1-t^2}{1+t^2}$ and $\sin \theta =\frac{2t}{1+t^2}$ to rewrite the expression inside the integral

$\begin{array}{rcl}\frac{1}{2+2\cos \theta +\sin \theta }& =& \frac{1}{2+\frac{2\left(1-t^2\right)}{1+t^2}+\frac{2t}{1+t^2}}\\ & & \end{array}$

Multiply top and bottom by $\left(1+t^2\right)$ to simplify the result

$\begin{array}{rcl}\frac{1}{2+2\cos \theta +\sin \theta }& =& \frac{1}{2+\frac{2\left(1-t^2\right)}{1+t^2}+\frac{2t}{1+t^2}}\times \frac{1+t^2}{1+t^2}\\ & =& \frac{1+t^2}{2\left(1+t^2\right)+2\left(1-t^2\right)+2t}\\ & =& \frac{1+t^2}{2+2t^2+2-2t^2+2t}\\ & =& \frac{1+t^2}{2\left(t+2\right)}\end{array}$

Change $d\theta$ into $\mathrm{d}t$ by differentiating $t=\tan \left(\frac{\theta }{2}\right)$

$\frac{\mathrm{d}t}{\mathrm{d}\theta }=\frac{1}{2}{\sec }^2\left(\frac{\theta }{2}\right)$

Use $1+{\tan }^{2}A\equiv {\sec }^{2}A$ to write it in terms of $t$

$\begin{array}{rcl}\frac{\mathrm{d}t}{\mathrm{d}\theta }& =& \frac{1}{2}\left(1+{\tan }^2\left(\frac{\theta }{2}\right)\right)\\ \frac{\mathrm{d}t}{\mathrm{d}\theta }& =& \frac{1}{2}\left(1+t^2\right)\end{array}$

Find $d\theta$ in terms of $\mathrm{d}t$

$\begin{array}{rcl}\frac{\mathrm{d}\theta }{\mathrm{d}t}& =& \frac{2}{1+t^2}\\ \mathrm{d}\theta & =& \frac{2}{1+t^2}\mathrm{d}t\end{array}$

Change the limits

$\begin{array}{l}\theta =\frac{\pi }{3}\Rightarrow t=\tan \left(\frac{\frac{\pi }{3}}{2}\right)=\tan \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{3}\\ \theta =\frac{\pi }{2}\Rightarrow t=\tan \left(\frac{\frac{\pi }{2}}{2}\right)=\tan \left(\frac{\pi }{4}\right)=1\end{array}$

Substitute the expression, $d\theta$ and new limits into the integral and simplify

$\begin{array}{rcl}{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2+2\cos \theta +\sin \theta }d\theta & =& {\int }_{\frac{\sqrt{3}}{3}}^1\frac{\cancel{1+t^2}}{\bcancel{2}\left(t+2\right)}\times \frac{\bcancel{2}}{\cancel{1+t^2}}\mathrm{d}t\\ & =& {\int }_{\frac{\sqrt{3}}{3}}^1\frac{1}{t+2}\mathrm{d}t\end{array}$

Integrate and substitute in the limits

$\begin{array}{rcl}{\int }_{\frac{\sqrt{3}}{3}}^1\frac{1}{t+2}\mathrm{d}t& =& {\left[\ln |t+2|\right]}_{\frac{\sqrt{3}}{3}}^1\\ & =& \ln 3-\ln \left(\frac{\sqrt{3}}{3}+2\right)\end{array}$

Use log laws and rationalising the denominator (or your calculator) to write out the final answer in the form $\ln \left(p+q\sqrt{3}\right)$

$\begin{array}{rcl}{\int }_{\frac{\sqrt{3}}{3}}^1\frac{1}{t+2}\mathrm{d}t& =& \ln \left(\frac{3}{{\displaystyle \frac{\sqrt{3}}{3}}+2}\right)\\ & =& \ln \left(\frac{9}{{\displaystyle \sqrt{3}+6}}\right)\\ & =& \ln \left(\frac{9}{{\displaystyle \sqrt{3}+6}}\times \frac{\sqrt{3}-6}{\sqrt{3}-6}\right)\\ & =& \ln \left(\frac{9\sqrt{3}-54}{{\displaystyle -33}}\right)\\ & =& \ln \left(\frac{18-3\sqrt{3}}{11}\right)\end{array}$

Write as $\ln \left(p+q\sqrt{3}\right)$

${\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2+2\cos \theta +\sin \theta }d\theta =\ln \left(\frac{18}{11}-\frac{3}{11}\sqrt{3}\right)$

Use the substitution $t=\tan \left(\frac{\theta }{2}\right)$ to determine the exact value of

${\int }_0^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta$

Answer:

The substitution $t=\tan \left(\frac{\theta }{2}\right)$ is undefined at $\theta =\pi$ which lies between the two limits

$0<\pi \mathit{<}\frac{3\pi }{2}$

Split the integral into two separate integrals, either side of $\theta =\pi$

${\int }_0^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta =\underset{a\to {\pi }^{-}}{\lim }{\int }_0^a\frac{1}{2+\cos \theta }d\theta +\underset{b\to {\pi }^{+}}{\lim }{\int }_b^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta$

Use the t-formula $\cos \theta =\frac{1-t^2}{1+t^2}$ to rewrite the expression inside the integral

$\begin{array}{rcl}\frac{1}{2+\cos \theta }& =& \frac{1}{2+\frac{1-t^2}{1+t^2}}\end{array}$

Multiply top and bottom by $\left(1+t^2\right)$ to simplify the result

$\begin{array}{rcl}\frac{1}{2+\cos \theta }& =& \frac{1}{2+\frac{1-t^2}{1+t^2}}\times \frac{1+t^2}{1+t^2}\\ & =& \frac{1+t^2}{2\left(1+t^2\right)+\left(1-t^2\right)}\\ & =& \frac{1+t^2}{2+2t^2+1-t^2}\\ & =& \frac{1+t^2}{t^2+3}\end{array}$

Change $d\theta$ into $\mathrm{d}t$ by differentiating $t=\tan \left(\frac{\theta }{2}\right)$

$\frac{\mathrm{d}t}{\mathrm{d}\theta }=\frac{1}{2}{\sec }^2\left(\frac{\theta }{2}\right)$

Use $1+{\tan }^{2}A\equiv {\sec }^{2}A$ to write it in terms of $t$

$\begin{array}{rcl}\frac{\mathrm{d}t}{\mathrm{d}\theta }& =& \frac{1}{2}\left(1+{\tan }^2\left(\frac{\theta }{2}\right)\right)\\ \frac{\mathrm{d}t}{\mathrm{d}\theta }& =& \frac{1}{2}\left(1+t^2\right)\end{array}$

Find $d\theta$ in terms of $\mathrm{d}t$

$\begin{array}{rcl}\frac{\mathrm{d}\theta }{\mathrm{d}t}& =& \frac{2}{1+t^2}\\ \mathrm{d}\theta & =& \frac{2}{1+t^2}\mathrm{d}t\end{array}$

Change the four limits

$\begin{array}{l}\theta =0\Rightarrow t=\tan \left(\frac{0}{2}\right)=\tan \left(0\right)=0\\ \theta =a\to {\pi }^{-}\Rightarrow \underset{a\to {\pi }^{-}}{\lim }\tan \left(\frac{a}{2}\right)=\infty \\ \theta =b\to {\pi }^{+}\Rightarrow \underset{b\to {\pi }^{+}}{\lim }\tan \left(\frac{b}{2}\right)=-\infty \\ \theta =\frac{3\pi }{2}\Rightarrow t=\tan \left(\frac{\frac{3\pi }{2}}{2}\right)=\tan \left(\frac{3\pi }{4}\right)=-1\end{array}$

Substitute the expression, $d\theta$ and new limits into the integrals and simplify

$\begin{array}{rcl}{\int }_0^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta & =& {\int }_0^{\infty }\frac{\cancel{1+t^2}}{t^2+3}\times \frac{2}{\cancel{1+t^2}}\mathrm{d}t+{\int }_{-\infty }^{-1}\frac{\cancel{1+t^2}}{t^2+3}\times \frac{2}{\cancel{1+t^2}}\mathrm{d}t\\ & =& 2{\int }_0^{\infty }\frac{1}{t^2+3}\mathrm{d}t+2{\int }_{-\infty }^{-1}\frac{1}{t^2+3}\mathrm{d}t\end{array}$

Integrate $\frac{1}{t^2+3}$, e.g. using that $\frac{1}{a^2+x^2}$ integrates to $\frac{1}{a}\arctan \left(\frac{x}{a}\right)$ (from the formulae booklet)

$\int \frac{1}{t^2+3}\mathrm{d}t=\frac{1}{\sqrt{3}}\arctan \left(\frac{t}{\sqrt{3}}\right)+c$

Substitute this into the working above

$\begin{array}{rcl}{\int }_0^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta & =& \frac{2}{\sqrt{3}}{\left[\arctan \left(\frac{t}{\sqrt{3}}\right)\right]}_0^{\infty }+\frac{2}{\sqrt{3}}{\left[\arctan \left(\frac{t}{\sqrt{3}}\right)\right]}_{-\infty }^{-1}\end{array}$

Find the value of the first integral (using the idea that $\arctan (\infty )=\frac{\pi }{2}$)

$\begin{array}{rcl}\frac{2}{\sqrt{3}}{\left[\arctan \left(\frac{t}{\sqrt{3}}\right)\right]}_0^{\infty }& =& \frac{2}{\sqrt{3}}\left(\underset{t\to \infty }{\lim }\arctan \left(\frac{t}{\sqrt{3}}\right)-\arctan \left(0\right)\right)\\ & =& \frac{2}{\sqrt{3}}\left(\frac{\pi }{2}-0\right)\\ & =& \frac{\pi \sqrt{3}}{3}\end{array}$

Find the value of the second integral (using the idea that $\arctan \left(-\infty \right)=-\frac{\pi }{2}$)

$\begin{array}{rcl}\frac{2}{\sqrt{3}}{\left[\arctan \left(\frac{t}{\sqrt{3}}\right)\right]}_{-\infty }^{-1}& =& \frac{2}{\sqrt{3}}\left(\arctan \left(\frac{-1}{\sqrt{3}}\right)-\underset{t\to -\infty }{\lim }\arctan \left(\frac{t}{\sqrt{3}}\right)\right)\\ & =& \frac{2}{\sqrt{3}}\left(-\frac{\pi }{6}-\left(-\frac{\pi }{2}\right)\right)\\ & =& \frac{2\pi \sqrt{3}}{9}\end{array}$

Add the answers together to give the final answer

$\begin{array}{rcl}{\int }_0^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta & =& \frac{\pi \sqrt{3}}{3}+\frac{2\pi \sqrt{3}}{9}\end{array}$

$\begin{array}{rcl}{\int }_0^{\frac{3\pi }{2}}\frac{1}{2+\cos \theta }d\theta & =& \frac{5\pi \sqrt{3}}{9}\end{array}$