Question

Calculate $\int {\dfrac{{\sqrt {\cos 2x} }}{{\cos x}}} \,dx$.

Answer 1

Here in this question we have to integrate the given function, since the function involves the trigonometric function. We use the trigonometric formulas which relate to the question and by substitution method whenever it is necessary we are solving the given function.

Complete step by step answer:
In an integration we come across two kinds of integral namely, definite integral and indefinite integral.Now consider the given question.
$\Rightarrow \int {\dfrac{{\sqrt {\cos 2x} }}{{\cos x}}} \,dx$
This is an indefinite integral.
Taking the square root both numerator and denominator we have
$\Rightarrow \int {\sqrt {\dfrac{{\cos 2x}}{{{{\cos }^2}x}}} } \,dx$
By the double angle formula of trigonometry we have $\cos 2x = 2{\cos ^2}x - 1$
$\Rightarrow \int {\sqrt {\dfrac{{2{{\cos }^2}x - 1}}{{{{\cos }^2}x}}} } \,dx$
On simplifying we get
$\Rightarrow \int {\sqrt {2 - {{\sec }^2}x} } \,dx$
By the trigonometric identity we know that $1 + {\tan ^2}x = {\sec ^2}x$
$\Rightarrow \int {\sqrt {2 - (1 + {{\tan }^2}x)} } \,dx$
On considering the sign conventions the above term is written as
$\Rightarrow \int {\sqrt {2 - 1 - {{\tan }^2}x} } \,dx$
$\Rightarrow \int {\sqrt {1 - {{\tan }^2}x} } \,dx$----(1)

Now substitute $\tan x = \sin y$----(2), on differentiating this we get ${\sec ^2}x\,dx = \cos y\,dy$
$\Rightarrow dx = \dfrac{{\cos y}}{{{{\sec }^2}x}}\,dy = \dfrac{{\cos y}}{{1 + {{\tan }^2}x}}\,dy = \dfrac{{\cos y}}{{1 + {{\sin }^2}y}}\,dy$ ----(3)
On substituting the equation (2) and equation (3) in the equation (1) we have
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}y} } \,.\dfrac{{\cos y}}{{1 + {{\sin }^2}y}}\,dy$
By the trigonometric identity we know that $\sqrt {1 - {{\sin }^2}y} = \cos y$
$\Rightarrow \int {\cos y} \,.\dfrac{{\cos y}}{{1 + {{\sin }^2}y}}\,dy$
$\Rightarrow \int {\dfrac{{{{\cos }^2}y}}{{1 + {{\sin }^2}y}}\,dy}$
In the numerator add 1 and subtract 1 we have
$\Rightarrow \int {\dfrac{{{{\cos }^2}y + 1 - 1}}{{1 + {{\sin }^2}y}}\,dy}$
By the trigonometric identity we know that $1 - {\sin ^2}y = {\cos ^2}y$
$\Rightarrow \int {\dfrac{{1 - {{\sin }^2}y + 1 - 1}}{{1 + {{\sin }^2}y}}\,dy}$

On simplifying the numerator term we have
$\Rightarrow \int {\dfrac{{2 - 1 - {{\sin }^2}y}}{{1 + {{\sin }^2}y}}\,dy}$
$\Rightarrow \int {\dfrac{{2 - (1 + {{\sin }^2}y)}}{{1 + {{\sin }^2}y}}\,dy}$
Taking the integral to each term
$\Rightarrow \int {\dfrac{2}{{1 + {{\sin }^2}y}}\,dy} + \int {\dfrac{{ - (1 + {{\sin }^2}y)}}{{1 + {{\sin }^2}y}}} \,dy$
$\Rightarrow \int {\dfrac{2}{{1 + {{\sin }^2}y}}\,dy} - \int {dy}$
On applying the integration to the second term
$\Rightarrow - y + \int {\dfrac{2}{{1 + {{\sin }^2}y}}\,dy}$
By the trigonometric identity we know that ${\sin ^2}y + {\cos ^2}y = 1$
$\Rightarrow - y + \int {\dfrac{2}{{{{\sin }^2}y + {{\cos }^2}y + {{\sin }^2}y}}\,dy}$
$\Rightarrow - y + \int {\dfrac{2}{{2{{\sin }^2}y + {{\cos }^2}y}}\,dy}$
Multiply both the numerator and denominator term by ${\sec ^2}y$
$\Rightarrow - y + \int {\dfrac{{2{{\sec }^2}y}}{{2{{\sin }^2}y{{\sec }^2}y + {{\cos }^2}y{{\sec }^2}y}}\,dy}$

By using the definition of trigonometric ratio we have
$\Rightarrow - y + \int {\dfrac{{2{{\sec }^2}y}}{{2{{\tan }^2}y + 1}}\,dy}$
$\Rightarrow - y + \int {\dfrac{{\sqrt 2 .\sqrt 2 {{\sec }^2}y}}{{1 + \left( {\sqrt 2 \tan y} \right)}}\,dy}$------(4)
Substitute $\sqrt 2 \tan y = u$----(5), on differentiating we have
$\Rightarrow \sqrt 2 {\sec ^2}y\,dy = du$---(6)
On substituting the equation (6) and equation (5) in the equation (4)
$\Rightarrow - y + \int {\dfrac{{\sqrt 2 \,du}}{{1 + {u^2}}}\,}$
On applying the integration we get
$\Rightarrow - y + \sqrt 2 {\tan ^{ - 1}}(u) + c$
On substituting the value of u we have
$\Rightarrow - y + \sqrt 2 {\tan ^{ - 1}}(2\tan y) + c$
On substituting the value of y we get
$\therefore - {\sin ^{ - 1}}(\tan x) + \sqrt 2 {\tan ^{ - 1}}(2\tan {\sin ^{ - 1}}(\tan x)) + c$

Note: When we are integrating the function by substitution method, after applying the integration we have to substitute the value which was considered. On substituting the function will be in the simplest form where we can integrate easily. Sometimes we need to substitute the terms more than once where it is necessary.