Question

Find anti derivative or integral of$\sqrt {\cos x}$?

Answer 1

Here you have to use trigonometric identity $\,{\sin ^2}x = 1 - {\cos ^2}x\,dx$ and you have to use the variable changing method i.e. an variable is replaced by another variable by simple assumption and then in last it is again replaced with the original one. It is very simple to use, not a very tough task.

Complete step by step answer:
Firstly we use the trigonometric identity $\,{\sin ^2}x = 1 - {\cos ^2}x\,dx$ and replace in our question as:
$\Rightarrow \cos x = \sqrt {1 - 2{{\sin }^2}x(\dfrac{1}{2})}$
Our question now transforms as: $\int {\sqrt {1 - 2{{\sin }^2}x(\dfrac{1}{2})} } dx$
Let $u = \dfrac{x}{2}$then $du = \dfrac{1}{2}dx$
Now, $I = 2\int {\sqrt {1 - 2{{\sin }^2}u} } \,du$
By using incomplete elliptic type of integration say “Y’, it states that:
$\Rightarrow Y(\dfrac{z}{{{k^2}}}) = \int\limits_0^z {\sqrt {1 - {k^2}{{\sin }^2}(\theta )} } = 2Y(\dfrac{z}{2}) + C$
So by using this our solution came as,
$\Rightarrow 2Y(\dfrac{x}{2}) + C$ or $2\sin (\dfrac{x}{2}) + C$
Now substituting $\sin x = \sqrt {1 - 2\operatorname{s} {{\cos }^2}x(\dfrac{1}{2})}$
We get,
$\Rightarrow 2\sqrt {1 - 2\operatorname{s} {{\cos }^2}x \times \dfrac{x}{2}(\dfrac{1}{2})}$ or $2\sqrt {1 - 2\operatorname{s} {{\cos }^2}{x^2}(\dfrac{1}{4})}$

Additional Information: Integration is quite complicated to solve, because for every question there are certain fixed steps you have to take or otherwise you will just keep on solving and results will not come or inappropriate that is the simplest result will not come.

Note: In the above question you can use some else method like try to solve with by parts method , it will also give you the same result but take almost triple steps what we see here, so for this type of question elliptic method should be used for fast answer. In case you forget this formula then you can simply use parts by assuming product “one into root cosx”and now you can solve with parts method, only the thing is you have to use parts a couple of times.