Question

Find the value of \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\tan x} }}} \].
A \[\dfrac{\pi }{{12}}\]
B \[\dfrac{\pi }{2}\]
C \[\dfrac{\pi }{6}\]
D \[\dfrac{\pi }{4}\]

Answer 1

Hint: o find the value of the integration first concert the function in integrable form. Then integrate the function and apply the limits.

Formula used: \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
\[\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} \]
\[\begin{array}{l}\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \\\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \end{array}\]

Complete step by step solution: The given integration \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\tan x} }}} \].
Consider, \[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\tan x} }}} \]
Substitute \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] in \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\tan x} }}} \] as follows.
\[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\dfrac{{\sin x}}{{\cos x}}} }}} \]
Now, separate square roots.
\[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \dfrac{{\sqrt {\sin x} }}{{\sqrt {\cos x} }}}}} \]
Now, simplify the denominator.
\[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{\dfrac{{\sqrt {\cos x} + \sqrt {\sin x} }}{{\sqrt {\cos x} }}}}} \]
Take the inverse of the denominator.
\[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos x} dx}}{{\sqrt {\cos x} + \sqrt {\sin x} }}} \] …(1)
Apply the property to simplify the function.
 \[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos \left( {\dfrac{\pi }{6} + \dfrac{\pi }{3} - x} \right)} dx}}{{\sqrt {\cos \left( {\dfrac{\pi }{6} + \dfrac{\pi }{3} - x} \right)} + \sqrt {\sin \left( {\dfrac{\pi }{6} + \dfrac{\pi }{3} - x} \right)} }}} \]
Further write in the following way.
\[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos \left( {\dfrac{\pi }{2} - x} \right)} dx}}{{\sqrt {\cos \left( {\dfrac{\pi }{2} - x} \right)} + \sqrt {\sin \left( {\dfrac{\pi }{2} - x} \right)} }}} \]
Simplify the function.
\[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\sin x} dx}}{{\sqrt {\sin x} + \sqrt {\cos x} }}} \] …(2)
Add equation (1) and equation (2)
\[I + I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos x} dx}}{{\sqrt {\cos x} + \sqrt {\sin x} }}} + \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\sin x} dx}}{{\sqrt {\sin x} + \sqrt {\cos x} }}} \]
Now, the equation becomes as follows.
\[\begin{array}{l}2I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos x} + \sqrt {\sin x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}dx} \\2I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {1dx} \end{array}\]
Evaluate the equation.
\[2I = \left[ x \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}\]
Apply the limit. Subtract lower limit from upper limit.
\[2I = \left[ {\dfrac{\pi }{3} - \dfrac{\pi }{6}} \right]\]
Simplify the bracket.
\[\begin{array}{l}2I = \left[ {\dfrac{{2\pi }}{6} - \dfrac{\pi }{6}} \right]\\2I = \dfrac{\pi }{6}\end{array}\]
Divide the equation by \[2\] on both sides.
\[I = \dfrac{\pi }{{12}}\]
Hence the value of \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\tan x} }}} \]is \[\dfrac{\pi }{{12}}\].

Thus, Option (A) is correct.

Note: The common mistake made by students is taking integration of \[\sqrt {\tan x} \] and substituting limits to find value of the integration which is the wrong method of finding a solution.