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!