Question

The integral \[\int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \sqrt[3]{{\tan 2x}}}}} \] is equal to
(a) \[\dfrac{\pi }{{18}}\]
(b) \[\dfrac{\pi }{3}\]
(c) \[\dfrac{\pi }{{12}}\]
(d) \[\dfrac{\pi }{6}\]

Answer 1

Hint:
Here in the question, we are required to find the value of the definite integral. We will convert the tangent to sine and cosine and simplify the expression to get an equation. Then, we will use the property of definite integral and simplify it to get another equation. We will then add the two equations and simplify the expression. Then we will substitute the limits to get the value of the integral.
Formula Used: The property of definite integrals \[\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} \]. The sine and cosine of two complementary angles is given by \[\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x\] and \[\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x\].

Complete step by step solution:
We will use the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and the property of definite integrals \[\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} \] to get the value of the integral.
First, let \[I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \sqrt[3]{{\tan 2x}}}}} \].
We know that the tangent of an angle is given by \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
Therefore, we get
\[\tan 2x = \dfrac{{\sin 2x}}{{\cos 2x}}\]
Substituting \[\tan 2x = \dfrac{{\sin 2x}}{{\cos 2x}}\] in the equation \[I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \sqrt[3]{{\tan 2x}}}}} \], we get
\[I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \sqrt[3]{{\dfrac{{\sin 2x}}{{\cos 2x}}}}}}} \]
Simplifying the denominator, we get
\[\begin{array}{l} \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \dfrac{{\sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\cos 2x}}}}}}} \\ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{\dfrac{{\sqrt[3]{{\cos 2x}} + \sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\cos 2x}}}}}}} \end{array}\]
Multiplying the numerator and denominator by \[\sqrt[3]{{\cos 2x}}\], and rewriting the equation, we get
\[\begin{array}{l} \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2x}}}}{{\sqrt[3]{{\cos 2x}} + \sqrt[3]{{\sin 2x}}}}} dx\\ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx \ldots \ldots \ldots \left( 1 \right)\end{array}\]
Now, using the property of definite integrals \[\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} \], we will simplify the above expression.
Therefore, we get
\[ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2\left( {\dfrac{\pi }{{24}} + \dfrac{{5\pi }}{{24}} - x} \right)}}}}{{\sqrt[3]{{\sin 2\left( {\dfrac{\pi }{{24}} + \dfrac{{5\pi }}{{24}} - x} \right)}} + \sqrt[3]{{\cos 2\left( {\dfrac{\pi }{{24}} + \dfrac{{5\pi }}{{24}} - x} \right)}}}}} dx\]
Adding and subtracting the terms, we get
\[\begin{array}{l} \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2\left( {\dfrac{{6\pi }}{{24}} - x} \right)}}}}{{\sqrt[3]{{\sin 2\left( {\dfrac{{6\pi }}{{24}} - x} \right)}} + \sqrt[3]{{\cos 2\left( {\dfrac{{6\pi }}{{24}} - x} \right)}}}}} dx\\ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2\left( {\dfrac{\pi }{4} - x} \right)}}}}{{\sqrt[3]{{\sin 2\left( {\dfrac{\pi }{4} - x} \right)}} + \sqrt[3]{{\cos 2\left( {\dfrac{\pi }{4} - x} \right)}}}}} dx\end{array}\]
Multiplying \[\left( {\dfrac{\pi }{4} - x} \right)\] by 2 in the equation, we get
\[ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos \left( {\dfrac{\pi }{2} - 2x} \right)}}}}{{\sqrt[3]{{\sin \left( {\dfrac{\pi }{2} - 2x} \right)}} + \sqrt[3]{{\cos \left( {\dfrac{\pi }{2} - 2x} \right)}}}}} dx\]
Now, we know that sine and cosine of two complementary angles are equal, that is \[\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x\] and \[\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x\].
Therefore, we get \[\cos \left( {\dfrac{\pi }{2} - 2x} \right) = \sin 2x\] and \[\sin \left( {\dfrac{\pi }{2} - 2x} \right) = \cos 2x\].
Substituting \[\cos \left( {\dfrac{\pi }{2} - 2x} \right) = \sin 2x\] and \[\sin \left( {\dfrac{\pi }{2} - 2x} \right) = \cos 2x\] in the equation \[I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos \left( {\dfrac{\pi }{2} - 2x} \right)}}}}{{\sqrt[3]{{\sin \left( {\dfrac{\pi }{2} - 2x} \right)}} + \sqrt[3]{{\cos \left( {\dfrac{\pi }{2} - 2x} \right)}}}}} dx\], we get
\[ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\cos 2x}} + \sqrt[3]{{\sin 2x}}}}} dx\]
Rewriting the equation, we get
\[ \Rightarrow I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx \ldots \ldots \ldots \left( 2 \right)\]
Now, adding equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[\begin{array}{l}I + I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx + \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx\\ \Rightarrow 2I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx + \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx\end{array}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 2I = \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}{{\sqrt[3]{{\cos 2x}} + \sqrt[3]{{\sin 2x}}}}} dx\\ \Rightarrow 2I = \int\limits_{\pi /24}^{5\pi /24} {\left( 1 \right)} dx\end{array}\]
Also, we know that the integral of a constant \[\int {\left( 1 \right)} dx\] is \[x\].
Thus, we get
\[ \Rightarrow 2I = \left. x \right|_{\pi /24}^{5\pi /24}\]
Substituting the limits and simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 2I = \dfrac{{5\pi }}{{24}} - \dfrac{\pi }{{24}}\\ \Rightarrow 2I = \dfrac{{4\pi }}{{24}}\\ \Rightarrow 2I = \dfrac{\pi }{6}\end{array}\]
Dividing both sides by 2, we get
\[\begin{array}{l} \Rightarrow \dfrac{{2I}}{2} = \dfrac{\pi }{{2 \times 6}}\\ \Rightarrow I = \dfrac{\pi }{{12}}\\ \Rightarrow \int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \sqrt[3]{{\tan 2x}}}}} = \dfrac{\pi }{{12}}\end{array}\]
Therefore, the value of the definite integral \[\int\limits_{\pi /24}^{5\pi /24} {\dfrac{{dx}}{{1 + \sqrt[3]{{\tan 2x}}}}} \] is \[\dfrac{\pi }{{12}}\].

Hence, option (c) is correct.

Note:
We have used the sum of two integrals to add the integrals \[\int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\cos 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx\] and \[\int\limits_{\pi /24}^{5\pi /24} {\dfrac{{\sqrt[3]{{\sin 2x}}}}{{\sqrt[3]{{\sin 2x}} + \sqrt[3]{{\cos 2x}}}}} dx\]. The sum of two integrals \[f\left( x \right)\] and \[g\left( x \right)\] can be written as \[\int {f\left( x \right)} dx + \int {g\left( x \right)} dx = \int {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx\].