# Integration by Substitution with $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ The angle was halved

I've been encountering the problem of the below integration.

$$\int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } \tag{1}$$

The official description states that the above integration formula can be calculated using substitution of integration.

$$t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right)$$

$$\therefore ~~ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } = \frac{ 2\pi }{ \sqrt{ R ^{2} -r ^{2} } }$$

Currently I can't derive the above RHS.

I think firstly find out the form of result of calculations of indefinite integral of eqn1 is wiser way.

What I tried so far are as below.

$$t= \tan^{}\left( \frac{ \theta_{} }{ 2 } \right)$$

$$\frac{ \theta_{} }{ 2 }= \tan^{-1} \left( t \right) ~~ \leftarrow~~ \text{Thought that this approach won't work}$$

$$\frac{ dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) \cdot \frac{1}{2}$$

$$\frac{ 2dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right)$$

$$\frac{ d\theta }{ 2 dt } =\sec^{-2}\left( \frac{ \theta_{} }{ 2 } \right)$$

$$\frac{ d\theta }{ 2 dt } = \left( \sec^{}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2}$$

$$\frac{ d\theta }{ 2 dt } = \left( \cos^{-1}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2}$$

$$\frac{ d\theta }{ 2 dt } = \cos^{2}\left( \frac{ \theta_{} }{ 2 } \right)$$

$$d\theta = 2 dt \cdot \cos^{2}\left(\frac{ \theta_{} }{ 2 } \right)$$

First things to first, the equation1 has $$~ \cos^{}\left(\theta_{} \right) ~$$ however how can I handle $$~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$$ ??

The goal with the $$t=\tan\frac{\theta}{2}$$ substitution is to find all trig functions in terms of $$t$$. Of course with any integral substitution, the first thing is to find $$\frac{dt}{d\theta}$$. $$\frac{dt}{d\theta}=\frac{1}{2}\sec^2\frac{\theta}{2}$$ $$\frac{dt}{d\theta}=\frac{t^2+1}{2}$$

Since we also have a $$\cos\theta$$ term in our integrand, we need to find $$\cos\theta$$ in terms of $$t$$. We start with $$\cos\theta=2\cos^2\frac{\theta}{2}-1$$ $$\cos\theta=\frac{2}{\sec^2\frac{\theta}{2}}-1$$ $$\cos\theta=\frac{2}{t^2+1}-1$$ $$\cos\theta=\frac{1-t^2}{t^2+1}$$

As for the bounds of the resulting integral, we can split the bounds in half, apply the substitution, and then merge the two bounds. This gives us a bounds of $$t\in (-\infty,\infty)$$, providing that both halves are finite. Our integral is now $$\int_{-\infty}^\infty \frac{2}{t^2+1}\cdot\frac{1}{R+\frac{1-t^2}{t^2+1}r}\, dt$$ $$=\int_{-\infty}^\infty 2\cdot\frac{1}{(R-r)t^2+(R+r)}\, dt$$ $$=\frac{2}{R+r}\int_{-\infty}^\infty \frac{dt}{\frac{R-r}{R+r}t^2+1}$$ $$=\frac{2}{\sqrt{R^2-r^2}}\left[\tan^{-1} \left(\sqrt{\frac{R-r}{R+r}}t\right)\right]_{-\infty}^\infty$$ As long as $$R>r$$, then we can see that $$\lim_{t\to\infty} \tan^{-1} \left(\sqrt{\frac{R-r}{R+r}}t\right)=\frac{\pi}{2}$$ $$\lim_{t\to -\infty} \tan^{-1} \left(\sqrt{\frac{R-r}{R+r}}t\right)=-\frac{\pi}{2}$$ Hence, our integral evaluates to $$=\boxed{\frac{2\pi}{\sqrt{R^2-r^2}}}$$

Hints:

1. The function is even, relative to the oy axis $$\Rightarrow$$ we can change the limits of integration $$2\int_{0}^{\pi} \dfrac{d \theta}{r(\frac{R}{r}+\cos(\theta))}$$
2. Replacement $$\theta =\arccos(t)$$ Then:

$$d\theta = -\frac{dt}{\sqrt{1-t^{2}}}$$ $$\cos(\arccos(t))=t$$

The integral is simplified to the form:

$$-2\int_{0}^{\pi}\frac{\frac{dt}{\sqrt{1-t^{2}}}}{r(\frac{R}{r}+t)}$$

Finally, you can use the Chebyshev substitution!

Try to solve the problem using the following two equations: $$$$\cos(2x)= 2\cos^2(x)-1, \\ \cos^2(x)= \frac{1}{1+ \tan^2(x)}.$$$$

It would be better if you would work on the problem by yourself now, however, I'll write a detailed answer below this line. If you don't want to read it then don't scroll further.

We want to calculate $$\int _0^{2\pi}\frac{d\theta}{R+ r\cos\theta}.$$ Set $$t= \tan\left(\frac\theta2 \right),$$ then $$\theta= 2\arctan(t)$$, and hence $$d \theta= \frac{2}{1+t^2}d t.$$ Moreover, $$\cos\theta= \cos(2\arctan t)= 2\cos^2(\arctan t)-1 = \frac{2}{1+ \tan^2(\arctan t)}-1= \frac{2}{1+t^2}-1= \frac{1-t^2}{1+t^2}.$$

Putting these together we have $$$$\int _0^{2\pi}\frac{d\theta}{R+ r\cos\theta}= 2\int _0^{\pi}\frac{d\theta}{R+ r\cos\theta}= 2\int_0^{+\infty} \frac{\frac{2}{1+t^2}}{R+ r\frac{1-t^2}{1+t^2}} dt= 2\int_0^{\infty} \frac{2}{R(1+t^2)+r (1-t^2)}dt= 4\int_0^\infty\frac{dt}{(R+r)+ t^2(R-r)}= \frac{4}{R+r}\int_0^\infty\frac{dt}{1+ \frac{R-r}{R+r}t^2}= \frac{4}{\sqrt{R+r}\sqrt{R-r}}\int_0^\infty \frac{du}{1+u^2}= \frac{4}{\sqrt{R^2-r^2}}[\arctan u]\Big|_0^{+\infty}= \frac{2\pi}{\sqrt{R^2-r^2}},$$$$ where we used the substitution $$u= \frac{\sqrt{R-r}}{\sqrt{R+r}}t$$.

Note that: $$\cos(\theta)=\cos(2\frac{\theta}2)=\cos^2(\frac{\theta}2)-\sin^2(\frac{\theta}2)=\frac{\cos^2(\frac{\theta}2)-\sin^2(\frac{\theta}2)}{\cos^2(\frac{\theta}2)+\sin^2(\frac{\theta}2)}$$ Simplify by $$\displaystyle{\cos^2(\frac{\theta}2)}$$: $$\cos(\theta)=\displaystyle{\frac{1-\tan^2(\frac{\theta}2)}{1+\tan^2(\frac{\theta}2)}}$$.

With the above, the integrand becomes: $$\displaystyle{\frac{1}{R+r\cdot\frac{1-\tan^2(\frac{\theta}2)}{1+\tan^2(\frac{\theta}2)}}=\frac{1+\tan^2(\frac{\theta}2)}{R(1+\tan^2(\frac{\theta}2))+r(1-\tan^2(\frac{\theta}2))}}$$

Substitute $$t=\tan(\frac{\theta}2)$$. It follows that $$dt=\frac12(1+\tan^2(\frac{\theta}2))d\theta$$ and the integral becomes: $$2\cdot2\int_0^\infty\frac{1}{R(1+t^2)+r(1-t^2)}dt$$ which you should be able to easily solve.