What is the value of the integral \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \sqrt {\tan x} }}} dx\]?
A. \[\dfrac{\pi }{2}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{6}\]
D. 1

Hint: Here, a definite integral is given. First, simplify the given integral by using the basic trigonometric ratio. Then, simplify the integral by using the integral rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] and trigonometric identities. After that, add this integral with the given integral and simplify it. Then, solve the integral using the integration formulas. In the end, apply the upper and lower limits to get the value of the given integral.



Formula Used:\[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]
Integration rule: \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
The sum rule of integration: \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\]
\[\int\limits_a^b n dx = \left[ {nx} \right]_a^b\], where \[n\] is a number.



Complete step by step solution:The given definite integral is \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \sqrt {\tan x} }}} dx\].
Let consider,
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \sqrt {\tan x} }}} dx\]
Simplify the equation by using the trigonometric ratio \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] .
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \sqrt {\dfrac{{\sin x}}{{\cos x}}} }}} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{\dfrac{{\sqrt {\cos x} + \sqrt {\sin x} }}{{\sqrt {\cos x} }}}}} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} dx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] on the right-hand side.
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos \left( {\dfrac{\pi }{2} - x} \right)} }}{{\sqrt {\cos \left( {\dfrac{\pi }{2} - x} \right)} + \sqrt {\sin \left( {\dfrac{\pi }{2} - x} \right)} }}} dx\]
Apply the properties of the trigonometric function \[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \] and \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \].
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx\] \[.....\left( 2 \right)\]
Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[ \Rightarrow I + I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} dx + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx\]
Apply the sum rule of integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\].
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }} + \dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{{\sqrt {\cos x} + \sqrt {\sin x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} 1 dx\]
\[ \Rightarrow I = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {dx} \]
Solve the integral by applying the integration formula \[\int\limits_a^b n dx = \left[ {nx} \right]_a^b\].
\[ \Rightarrow I = \dfrac{1}{2}\left[ x \right]_0^{\dfrac{\pi }{2}}\]
Apply the upper and lower limits.
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\dfrac{\pi }{2} - 0} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{4}\]
Therefore, \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{1 + \sqrt {\tan x} }}} dx = \dfrac{\pi }{4}\].



Option ‘B’ is correct



Note: The other way to solve the given integral is:
Directly apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\], without converting the trigonometric function \[\tan x\] in the basic trigonometric ratio \[\dfrac{{\sin x}}{{\cos x}}\]. After that, follow the above steps and calculate the value of the definite integral.