Question

Evaluate the integral $\int {\sqrt {\sec x} {\text{ }}} dx$

Answer 1

In order to solve this question first write what is given to us. This will give us a clear picture of what our approach should be. Here, in this question, we have to use the conversion of $\sec x$ into $\cos x$ and then use the basic formula $\cos x$ and thus we will get our required answer.

Complete step-by-step solution:
According to the given information we have $\int {\sqrt {\sec x} {\text{ }}} dx$
As we know that by the trigonometric identities i.e. $\sec x = \dfrac{1}{{\cos x}}$
Therefore, $\int {\sqrt {\sec x} {\text{ }}} dx = \int {\dfrac{1}{{\sqrt {\cos x} }}{\text{ }}} dx$ (equation 1)
Let $\cos x = p$
Differentiating both side with respect to x we get
$\dfrac{d}{{dx}}\cos x = \dfrac{d}{{dx}}p$
We know that $\dfrac{d}{{dx}}\cos x = - \sin x$
Therefore, $- \sin x = \dfrac{{dp}}{{dx}}$
$\Rightarrow$ $- \sin xdx = dp$
$\Rightarrow$ $dx = \dfrac{{dp}}{{ - \sin x}}$
We know that ${\sin ^2}x + {\cos ^2}x = 1$or $\sin x = \sqrt {1 - {{\cos }^2}x}$
Therefore, $dx = \dfrac{{dp}}{{ - \sqrt {1 - {{\cos }^2}x} }}$
Substituting the value of dx and cos x in equation 1 we get
$\int {\dfrac{1}{{\sqrt u }}{\text{ }}} \times \dfrac{{dp}}{{ - \sqrt {1 - {u^2}} }}$
Substituting the value of u in the above equation we get
$- \int {\dfrac{{dp}}{{\sqrt {\cos x\left( {1 - {{\cos }^2}x} \right)} }}{\text{ }}}$
The above function is an example of elliptical integration.

Note: In the above solution we used the term elliptical which can be explained as one of a number of related functions defined as the value of certain integrals. Now, you will notice the base before we move into its types. The one used in this question is an elliptic integral of the first kind.