Integral (2)

Problem:

$\int{\sqrt{\frac{4 - x}{4 + x}}dx}$

Solution:

This one is not easy, I am thinking about this identity

$\sqrt{\frac{1 - \sin \theta}{1 + \sin\theta}} = \frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}$

Therefore I let $x = 4 \sin \theta$, that gives $dx = 4 \cos \theta d\theta$, the integral becomes

$\begin{eqnarray*} &=& \int{\sqrt{\frac{4 - 4\sin\theta}{4 + 4\sin\theta}}4 \cos \theta d\theta} \\ &=& \int{2\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}4 \cos \theta d\theta} \\ &=& \int{2\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}4 \cos \theta d\theta} \\ &=& \int{2\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}4 \frac{1-\tan^2 \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}} d\theta} \end{eqnarray*}$

Now it is quite obvious that we should do the t - substitution

$\begin{eqnarray*} t &=& \tan \frac{\theta}{2} \\ dt &=& \sec^2 \frac{\theta}{2} \frac{1}{2} d \theta\\ dt &=& (1 + t^2) \frac{1}{2} d \theta \\ d \theta &=& \frac{2tdt}{1 + t^2} \\ \end{eqnarray*}$

$\begin{eqnarray*} & & \int{2\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}4 \frac{1-\tan^2 \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}} d\theta} \\ &=& \int{2\frac{1-t}{1+t}4 \frac{1-t^2}{1+t^2} \frac{2tdt}{1 + t^2}} \\ &=& 8\int{\frac{1}{1+t^2} - \frac{2t}{(1 + t^2)^2}dt} \end{eqnarray*}$

The first term can be solved by substitute $t = \tan \phi$, $dt = \sec^2 \phi d\phi$

$\begin{eqnarray*} & & \int{\frac{1}{1+t^2}dt} \\ &=&\int{\frac{1}{1+\tan^2 \phi}\sec^2 \phi d\phi} \\ &=&\int{\frac{1}{\sec^2 \phi}\sec^2 \phi d\phi} \\ &=&\int{d\phi} \\ &=&\phi \end{eqnarray*}$

The second term can be solved by substitute $u = 1 + t^2$, $du = 2tdt$

$\begin{eqnarray*} & & \int{\frac{2t}{(1 + t^2)^2}dt} \\ &=& \int{\frac{1}{u^2}du} \\ &=& \frac{-1}{u} \end{eqnarray*}$

Putting them together, we have

$\begin{eqnarray*} & & \int{\sqrt{\frac{4 - x}{4 + x}}dx} \\ &=& 8\int{\frac{1}{1+t^2} - \frac{2t}{(1 + t^2)^2}dt} \\ &=& 8(\phi + \frac{1}{u}) \\ &=& 8(\phi + \frac{1}{1 + t^2}) \end{eqnarray*}$

Now we have $t = \tan \phi, t = \tan \frac{\theta}{2}$, so we simply have $\phi = \frac{\theta}{2}$

$\begin{eqnarray*} & & 8(\frac{\theta}{2} + \frac{1}{1 + t^2}) \\ &=& 8(\frac{\theta}{2} + \frac{1}{1 + \tan^2 \frac{\theta}{2}}) \\ &=& 8(\frac{\theta}{2} + \frac{1}{\sec^2 \frac{\theta}{2}}) \\ &=& 8(\frac{\theta}{2} + \cos^2 \frac{\theta}{2}) \\ &=& 8(\frac{\theta}{2} + \frac{1 + \cos \theta}{2}) \\ &=& 4(\theta + 1 + \cos \theta) \end{eqnarray*}$

Now constant can be dropped for indefinite integral, and $x = 4 \sin \theta$, so finally we have

$\begin{eqnarray*} & & 4(\theta + \cos \theta) \\ &=& 4(\arcsin(\frac{x}{4}) + \cos \theta) \\ &=& 4(\arcsin(\frac{x}{4}) + \sqrt{1 - \sin^2 \theta}) \\ &=& 4(\arcsin(\frac{x}{4}) + \sqrt{1 - (\frac{x}{4})^2}) \end{eqnarray*}$

There we go, what a complicated problem!