Evaluate the integral \[\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^3}xdx} \]
A. $\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2}\log \left( {\sqrt 2 + 1} \right) $
B. $\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}\log \left( {\sqrt 2 + 1} \right) $
C. $2\sqrt 2 .\log \left( {\sqrt 2 } \right) $
D.$\dfrac{1}{{\sqrt 2 }}\log \left( {\sqrt 2 } \right) $

Hint: We had to only apply a reduction formula for \[\int {{{\sec }^n}xdx} \]. And after that we can put the limits. Reduction formula for the integration for \[\int {{{\sec }^n}xdx} \] is \[\int {{{\sec }^n}xdx = } \dfrac{{{{\sec }^{n - 1}}\left( x \right)\sin x}}{{n - 1}} + \dfrac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}\left( x \right)dx} \].

Complete step-by-step answer:
As we know that if we are given the trigonometric function with a power as integer then we can directly apply a reduction formula to find the integration value.
So, applying reduction formula to find the value of \[\int {{{\sec }^3}xdx} \]
\[ \Rightarrow \int {{{\sec }^n}xdx = } \dfrac{{{{\sec }^{n - 1}}\left( x \right)\sin x}}{{n - 1}} + \dfrac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}\left( x \right)dx} \] (1)
Putting the value of n = 3 in the above equation. \[ \Rightarrow \int {{{\sec }^3}xdx = } \dfrac{{{{\sec }^{3 - 1}}\left( x \right)\sin x}}{{3 - 1}} + \dfrac{{3 - 2}}{{3 - 1}}\int {{{\sec }^{3 - 2}}\left( x \right)dx} \]
Solving above equation.
\[ \Rightarrow \int {{{\sec }^3}xdx = } \dfrac{{{{\sec }^2}\left( x \right)\sin x}}{2} + \dfrac{1}{2}\int {\sec xdx} \] (2)
Now as we know that the integration of \[\sec x\] is \[\log \left| {\sec x + \tan x} \right|\].
So, \[\int {\sec xdx} = \log \left| {\sec x + \tan x} \right|\]
So, putting the value of \[\int {\sec xdx} \]in equation 2.
\[ \Rightarrow \int {{{\sec }^3}xdx = } \dfrac{{{{\sec }^2}\left( x \right)\sin x}}{2} + \dfrac{1}{2}\log \left| {\sec x + \tan x} \right|\]
Now applying limits from 0 to \[\dfrac{\pi }{4}\] to both the sides of the above equation.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^3}xdx} = \dfrac{1}{2}\left[ {{{\sec }^2}\left( x \right)\sin x} \right]_0^{\dfrac{\pi }{4}} + \dfrac{1}{2}\left[ {\log \left| {\sec x + \tan x} \right|} \right]_0^{\dfrac{\pi }{4}}\]
Now we had to put upper limits and lower limits in the above equation.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^3}xdx} = \dfrac{1}{2}\left[ {{{\sec }^2}\left( {\dfrac{\pi }{4}} \right)\sin \dfrac{\pi }{4} - {{\sec }^2}\left( 0 \right)\sin 0} \right] + \dfrac{1}{2}\left[ {\log \left| {\sec \dfrac{\pi }{4} + \tan \dfrac{\pi }{4}} \right| - \log \left| {\sec 0 + \tan 0} \right|} \right]\]
Now as we know that \[\sec \dfrac{\pi }{4} = \sqrt 2 \], \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\], \[\sec 0 = 1,\sin 0 = 0,\tan \dfrac{\pi }{4} = 1,\tan 0 = 0\] and according to logarithmic identities \[\log \left| a \right| - \log \left| b \right| = \log \left| {\dfrac{a}{b}} \right|\]
So, \[\int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^3}xdx} = \dfrac{1}{2}\left[ {{{\left( {\sqrt 2 } \right)}^2}\dfrac{1}{{\sqrt 2 }} - {{\left( 1 \right)}^2}0} \right] + \dfrac{1}{2}\left[ {\log \left| {\sqrt 2 + 1} \right| - \log \left| {1 + 0} \right|} \right]\]
Now solving the above equation.
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^3}xdx} = \dfrac{1}{2}\left[ {\sqrt 2 } \right] + \dfrac{1}{2}\left[ {\log \left| {\dfrac{{\sqrt 2 + 1}}{1}} \right|} \right]\]
\[ \Rightarrow \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^3}xdx} = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2}\log \left( {\sqrt 2 + 1} \right)\]
Hence, the correct option will be A.

Note:- Whenever we come up with this type of problem then there is also another way to find the solution. We can also apply by parts with the first term as \[u = \sec x\] and the second term as \[v = {\sec ^2}x\]. And then applying by-parts formula that is \[\int {uvdx} = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\left( {\int {vdx} } \right)dx} \right)} } \]. But the easiest and efficient way to find the value of the integral of type \[\int {{{\sec }^n}xdx} \] is by applying a reduction formula.