Question

What is the value of \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\cot x} }}} \]?
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{6}\]
C. \[\dfrac{\pi }{{12}}\]
D. \[\dfrac{\pi }{2}\]

Answer 1

Hint: First we will rewrite \[\cot x\] as ratio \[\cos x\] and \[\sin x\]. Then simplify it. After that we will apply the property of definite integral that is \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \] and add the original integration and the new integration that we get after applying the formula. Simplify the sum and integrate it to get the required answer.



Formula Used:Definite integral property:
\[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]
Complementary formula of trigonometry:
\[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]
Integration formula:
\[\int\limits_b^a {dx} = \left[ {a - b} \right]\]



Complete step by step solution:Given definite integral is \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\cot x} }}} \].
Assume that, \[I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\cot x} }}} \]
Now we replace \[\cot x\] by \[\dfrac{{\cos x}}{{\sin x}}\]:
\[ \Rightarrow I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\dfrac{{\cos x}}{{\sin x}}} }}} \]
Simplify the above expression:
\[ \Rightarrow I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{\dfrac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} }}}}} \]
\[ \Rightarrow I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\sin x} dx}}{{\sqrt {\sin x} + \sqrt {\cos x} }}} \] …….(i)
Now applying the property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]:
\[ \Rightarrow I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6} - x} \right)} dx}}{{\sqrt {\sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6} - x} \right)} + \sqrt {\cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6} - x} \right)} }}} \]
Add the like terms:
\[ \Rightarrow I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\sin \left( {\dfrac{\pi }{2} - x} \right)} dx}}{{\sqrt {\sin \left( {\dfrac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\dfrac{\pi }{2} - x} \right)} }}} \]
Now applying the complementary formula of trigonometry:
\[ \Rightarrow I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos x} dx}}{{\sqrt {\cos x} + \sqrt {\sin x} }}} \] …….(ii)
Now adding equation (i) and (ii)
\[I + I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\sin x} dx}}{{\sqrt {\sin x} + \sqrt {\cos x} }}} + \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt {\cos x} dx}}{{\sqrt {\cos x} + \sqrt {\sin x} }}} \]
\[ \Rightarrow 2I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\left( {\sqrt {\sin x} + \sqrt {\cos x} } \right)dx}}{{\left( {\sqrt {\sin x} + \sqrt {\cos x} } \right)}}} \]
Now cancel out common term from denominator and numerator:
\[ \Rightarrow 2I = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {1dx} \]
\[ \Rightarrow 2I = \left( {\dfrac{\pi }{3} - \dfrac{\pi }{6}} \right)\]
\[ \Rightarrow 2I = \dfrac{\pi }{3}\]
Divide both sides by 2:
\[ \Rightarrow I = \dfrac{\pi }{6}\]



Option ‘B’ is correct



Note: Students often make mistake to solve the given question. They apply the property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \] in \[\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{dx}}{{1 + \sqrt {\cot x} }}} \] and stuck in the step. First we have replace \[\cot x\] by \[\dfrac{{\cos x}}{{\sin x}}\] and then apply the definite integral property.