Question

What is the value of definite integral \[\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{2}{3}}}x}}{{{{\sin }^{\dfrac{2}{3}}}x + {{\cos }^{\dfrac{2}{3}}}x}}} dx\]?
A. \[\dfrac{\pi }{4}\]
B. \[\dfrac{\pi }{2}\]
C. \[\dfrac{{3\pi }}{4}\]
D. \[\pi \]

Answer 1

Hint: The given integration is a definite integration. We will apply definite integral property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \] to simplify it. Then add the given integration with the new form of the integration and solve it.



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



Complete step by step solution:Given definite integration is \[\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{2}{3}}}x}}{{{{\sin }^{\dfrac{2}{3}}}x + {{\cos }^{\dfrac{2}{3}}}x}}} dx\].
Assume that, \[I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{2}{3}}}x}}{{{{\sin }^{\dfrac{2}{3}}}x + {{\cos }^{\dfrac{2}{3}}}x}}} dx\] …..(i)
Now applying the property of definite integration:
\[I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{2}{3}}}\left( {\dfrac{\pi }{2} - x} \right)}}{{{{\sin }^{\dfrac{2}{3}}}\left( {\dfrac{\pi }{2} - x} \right) + {{\cos }^{\dfrac{2}{3}}}\left( {\dfrac{\pi }{2} - x} \right)}}} dx\]
Now applying the complementary formula of trigonometry:
\[I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^{\dfrac{2}{3}}}x}}{{{{\cos }^{\dfrac{2}{3}}}x + {{\sin }^{\dfrac{2}{3}}}x}}} dx\] …..(ii)
Adding equation (i) and (ii)
\[I + I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{2}{3}}}x}}{{{{\sin }^{\dfrac{2}{3}}}x + {{\cos }^{\dfrac{2}{3}}}x}}} dx + \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^{\dfrac{2}{3}}}x}}{{{{\cos }^{\dfrac{2}{3}}}x + {{\sin }^{\dfrac{2}{3}}}x}}} dx\]
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{2}{3}}}x + {{\cos }^{\dfrac{2}{3}}}x}}{{{{\sin }^{\dfrac{2}{3}}}x + {{\cos }^{\dfrac{2}{3}}}x}}} dx\]
Now cancel out common term from denominator and numerator:
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {dx} \]
Now integrate the right side:
\[ \Rightarrow 2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow 2I = \left[ {\dfrac{\pi }{2} - 0} \right]\]
Divide both sides by 2:
\[ \Rightarrow I = \dfrac{\pi }{4}\]



Option ‘A’ is correct


Note: Students often make mistakes to solve the definite integration. They divide denominator and numerator by \[{\cos ^{\dfrac{2}{3}}}x\] and use substitution method to solve it. But it is incorrect way. Here we need to apply the property of definite integral and solve it.