Problem 43:  The $t$ substitution ($✓$ $✓$) 2000 Paper II

Show that

$sin𝜃=\frac{2t}{1+{t}^{2}}\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}cos𝜃=\frac{1-{t}^{2}}{1+{t}^{2}}\phantom{\rule{2.77695pt}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{1+cos𝜃}{sin𝜃}=tan\left(\frac{1}{2}\pi -\frac{1}{2}𝜃\right)\phantom{\rule{2.77695pt}{0ex}},$

where $t=tan\left(\frac{1}{2}𝜃\right)\phantom{\rule{0.3em}{0ex}}$.

Let

$I={\int }_{0}^{\frac{1}{2}\pi }\frac{1}{1+cos\alpha sin𝜃}\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃\phantom{\rule{2.77695pt}{0ex}}.$

Use the substitution $t=tan\left(\frac{1}{2}𝜃\right)$ to show that, for $0<\alpha <\frac{1}{2}\pi \phantom{\rule{0.3em}{0ex}}$,

$I={\int }_{0}^{1}\frac{2}{{\left(t+cos\alpha \right)}^{2}+{sin}^{2}\alpha }\phantom{\rule{0.3em}{0ex}}\mathrm{d}t\phantom{\rule{2.77695pt}{0ex}}.$

By means of the further substitution $t+cos\alpha =sin\alpha tanu$ show that

$I=\frac{\alpha }{sin\alpha }\phantom{\rule{2.77695pt}{0ex}}.$

Deduce a similar result for

${\int }_{0}^{\frac{1}{2}\pi }\frac{1}{1+sin\alpha cos\varphi }\phantom{\rule{0.3em}{0ex}}\mathrm{d}\varphi \phantom{\rule{2.77695pt}{0ex}}.$

The ﬁrst of the two substitutions is familiarly known as the ‘$t$ substitution’. It would have been very standard fare 30 years ago, but it seems to have gone out of fashion now. The second of the two substitutions is the normal substitution for integrals with quadratic denominators.

For the last part, ‘deduce’ implies that you don’t have to do any further integration. Note that the variable in the integral is $\varphi$ instead of $𝜃$. Since it is a deﬁnite integral, it doesn’t matter what the variable is called; it could equally well have been called $𝜃$ as in the original integral $I\phantom{\rule{0.3em}{0ex}}$. The use of a different variable was just a kindness on the part of the examiner to indicate that you should be thinking about a change of variable.

The ‘similar result’ result that you deduce should include the conditions under which it is true. It is worth thinking about why the condition $0<\alpha <\frac{1}{2}\pi$ is required — or, indeed, if it is required.

Solution to problem 43

The three identities just require use of $cos𝜃={cos}^{2}\frac{1}{2}𝜃-{sin}^{2}\frac{1}{2}𝜃$ and $sin𝜃=2sin\frac{1}{2}𝜃cos\frac{1}{2}𝜃\phantom{\rule{0.3em}{0ex}}$. If you divide each by ${cos}^{2}\frac{1}{2}𝜃+{sin}^{2}\frac{1}{2}𝜃$ (i.e. by 1) the ﬁrst two identities drop out. Remember, for the last one, that $cotx=tan\left(\frac{1}{2}\pi -x\right)$.

For the ﬁrst change of variable, we have $\mathrm{d}t=\frac{1}{2}{sec}^{2}\frac{1}{2}𝜃\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃=\frac{1}{2}\left(1+{t}^{2}\right)\mathrm{d}𝜃$ and the new limits are $0$ and $1$, so

$\begin{array}{rcll}I={\int }_{0}^{\frac{1}{2}\pi }\frac{1}{1+cos\alpha sin𝜃}\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃& =& {\int }_{0}^{1}\frac{1}{1+cos\alpha \frac{2t}{1+{t}^{2}}}\phantom{\rule{2.77695pt}{0ex}}\frac{2}{1+{t}^{2}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}t& \text{}\\ & =& {\int }_{0}^{1}\frac{2}{1+2tcos\alpha +{t}^{2}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}t={\int }_{0}^{1}\frac{2}{{\left(t+cos\alpha \right)}^{2}+{sin}^{2}\alpha }\phantom{\rule{0.3em}{0ex}}\mathrm{d}t\phantom{\rule{2.77695pt}{0ex}}.& \text{}\end{array}$

For the second change of variable, we have $\mathrm{d}t=sin\alpha {sec}^{2}u\phantom{\rule{0.3em}{0ex}}\mathrm{d}u\phantom{\rule{0.3em}{0ex}}$. When $t=0\phantom{\rule{0.3em}{0ex}}$, $tanu=cot\alpha$ so $u=\frac{1}{2}\pi -\alpha \phantom{\rule{0.3em}{0ex}}$. When $t=1\phantom{\rule{0.3em}{0ex}}$, $sin\alpha tanu=1+cos\alpha$ so (after a bit of work with double-angle formulae) $u=\frac{1}{2}\pi -\frac{1}{2}\alpha \phantom{\rule{0.3em}{0ex}}$. Thus

$I={\int }_{\frac{1}{2}\pi -\alpha }^{\frac{1}{2}\pi -\frac{1}{2}\alpha }\frac{2}{{sin}^{2}\alpha \left(1+{tan}^{2}u\right)}\phantom{\rule{2.77695pt}{0ex}}sin\alpha {sec}^{2}u\mathrm{d}u={\int }_{\frac{1}{2}\pi -\alpha }^{\frac{1}{2}\pi -\frac{1}{2}\alpha }\frac{2}{sin\alpha }\phantom{\rule{0.3em}{0ex}}\mathrm{d}u=\frac{\alpha }{sin\alpha }\phantom{\rule{2.77695pt}{0ex}}.$

For the last part, we want to make a substitution that changes the cosine in the denominator to a sine. One possibility is to set $\varphi =\frac{1}{2}\pi -𝜃\phantom{\rule{0.3em}{0ex}}$. This will swap the limits but also introduces a minus sign since $\mathrm{d}𝜃=-\mathrm{d}\varphi \phantom{\rule{0.3em}{0ex}}$. Thus

$\frac{\alpha }{sin\alpha }=I=-{\int }_{\frac{\pi }{2}}^{0}\frac{1}{1+cos\alpha cos\varphi }\phantom{\rule{0.3em}{0ex}}\mathrm{d}\varphi ={\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+cos\alpha cos\varphi }\phantom{\rule{0.3em}{0ex}}\mathrm{d}\varphi \phantom{\rule{2.77695pt}{0ex}}.$

This is almost the integral we want: we still need to replace $cos\alpha$ in the denominator by $sin\alpha \phantom{\rule{0.3em}{0ex}}$. Remembering what we did a couple of lines back, we just replace $\alpha$ by $\frac{1}{2}\pi -\alpha$ in the integral and in the answer, giving

${\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+cos\alpha cos\varphi }\phantom{\rule{0.3em}{0ex}}\mathrm{d}\varphi =\frac{\frac{1}{2}\pi -\alpha }{cos\alpha }\phantom{\rule{2.77695pt}{0ex}}.$

If the original $\alpha$ satisﬁed $0<\alpha <\frac{1}{2}\pi$, the new $\alpha$ must satisfy $\frac{1}{2}\pi >\alpha >0$, which is the same.

Post-mortem

One reason for the restriction on $\alpha$ might be to prevent the denominator of the integrand being zero for some value of $𝜃\phantom{\rule{0.3em}{0ex}}$; a zero in the denominator usually means that the integral is undeﬁned. However, the only value of $\alpha$ for which $cos\alpha sin𝜃$ could possibly be as small as $-1$ (for $0\le 𝜃\le \frac{1}{2}\pi$) is $\pi \phantom{\rule{0.3em}{0ex}}$ (and of course $3\pi$, etc). From this point of view, we only need $\alpha \ne \pi \phantom{\rule{0.3em}{0ex}}$ (etc). There is a slight awkwardness in the answer when $\alpha =0\phantom{\rule{0.3em}{0ex}}$ but this can be overcome by taking limits:

$\underset{\alpha \to 0}{lim}\frac{\alpha }{sin\alpha }=1$

which you can easily verify is the correct value of the integral when $\alpha =0$.

You might therefore think that the restriction $0<\alpha <\frac{1}{2}\pi$ is superﬂuous. But here is a curious thing: increasing $\alpha$ by $2\pi$ does not change $I$ but does change the answer! If you work through the solution with this in mind you see that it can make a difference only when you work out ${tan}^{-1}u\phantom{\rule{0.3em}{0ex}}$, which by deﬁnition lies in the range $-\frac{1}{2}\pi$ to $\frac{1}{2}\pi \phantom{\rule{0.3em}{0ex}}$, and this is why we must have $-\frac{1}{2}\pi <\alpha \le \frac{1}{2}\pi$. We lose nothing by using instead $0\le \alpha \le \frac{1}{2}\pi$, since $\alpha \to -\alpha$ doesn’t change the integral. The strict inequalities ($<$ rather than $\le$) avoid trouble with the denominators when $\alpha =0$ in the ﬁrst integral or $\alpha =\frac{1}{2}\pi$ in the second.