Question

Calculate the value of $\int {\dfrac{{\sin xdx}}{{{{\cos }^2}x\sqrt {\cos 2x} }}}$
A) $C - \sqrt {1 - {{\tan }^2}x}$
B) $C - \sqrt {1 - {{\sec }^2}x}$
C) $C - \sqrt {1 - {{\cos }^2}x}$
D) $C - \sqrt {1 - {{\sin }^2}x}$

Answer 1

The method of substitution will be used in this question.
According to this method, one variable is substituted in the equation by another variable to simplify it.
Thus many functions which cannot be integrated easily, are reduced into standard forms so that their integration becomes easy.
Then simple integration formulas and concepts are applied to get the answer.

Complete step-by-step answer:
The given equation is$\int {\dfrac{{\sin x}}{{{{\cos }^2}x\sqrt {\cos 2x} }}dx}$
Converting the question in tan x form, $\dfrac{{\sin x}}{{\cos x}} = \tan x$ and cos x is reciprocal of sec x i.e. $\dfrac{1}{{\cos x}} = \sec x$
Substituting these value in the equation, we get
$\Rightarrow$ $\int {\dfrac{{\sin x}}{{{{\cos }^2}x\sqrt {\cos 2x} }}dx} = \int {\dfrac{{\tan x\sec x}}{{\sqrt {\cos 2x} }}dx}$ ………… (1)
We have the formula that converts the cos function in tan function using the trigonometric identities to find the integral,
$\sqrt {\cos 2x} = \sqrt {\dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}}$
Putting the value of $\sqrt {\cos 2x}$ in the equation, we further get
\Rightarrow$$$\int {\dfrac{{\tan x\sec x}}{{\sqrt {\cos 2x} }}dx} = \int {\dfrac{{\tan x\sec x}}{{\sqrt {\dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} }}} dx$$ Here we have, 1 + {\tan ^2}x = {\sec ^2}x$so$\sqrt {1 + {{\tan }^2}x} = \sec x$Substituting this value in the equation,$\Rightarrow$$$\int {\dfrac{{\tan x\sec x}}{{\sqrt {\dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} }}} dx = \int {\dfrac{{\tan x{{\sec }^2}x}}{{\sqrt {1 - {{\tan }^2}x} }}dx} $………… (2) Now, it is often important to guess what will be the useful substitution, so let${\tan ^2}x = t$$, then differentiating with respect to t
We make the following transformation while using the method of substitution.
We have,

$$2\tan x{\sec ^2}x = dt \\\ \tan x{\sec ^2}x = \dfrac{{dt}}{2} \\\$$

Substituting the value in equation (2), we get:
$\Rightarrow$$$\int {\dfrac{{\tan x{{\sec }^2}x}}{{\sqrt {1 - {{\tan }^2}x} }}dx} = \dfrac{1}{2}\int {\dfrac{{dt}}{{\sqrt {1 - t} }}} $…….. (3) Now, again assuming that$1 - t = u$, differentiating with respect to t, we get:$ - dt = du$$
Putting this value in equation (3), the given integral can be written as

$$\dfrac{1}{2}\int {\dfrac{{dt}}{{\sqrt {1 - t} }}} = \dfrac{1}{2}\int {\dfrac{{du}}{{\sqrt u }}} \\\ = \dfrac{1}{2}\int {{u^{\dfrac{{ - 1}}{2}}}} du \\\$$

Integrating with respect to u, we have

$$\dfrac{1}{2}\int {{u^{\dfrac{{ - 1}}{2}}}} du = \dfrac{1}{2} \times - 2\sqrt u + C \\\ = C - \sqrt u \\\$$

Where C is the constant, putting the value of u in above equation
$\Rightarrow C - \sqrt u = C - \sqrt {1 - t}$
And now putting the value of t we have the final answer,
$\Rightarrow C - \sqrt {1 - t} = C - \sqrt {1 - {{\tan }^2}x}$

So option (A) is the correct answer.

Note: When the integration involves some of the trigonometric functions, we use some identities to reduce the integral into the easiest possible form.
For cos2x there are 4 identities. So choose the identity that simplifies the whole equation.
Here, the whole equation is converted in tan and sec function.
$C$ in the final answer is a constant value.