Question

If \[\int\limits_0^\pi {xf\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx = k} \int\limits_0^{\dfrac{\pi }{2}} {f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \], then what is the value of \[k\]?
A. \[\dfrac{\pi }{2}\]
B. \[\pi \]
C. \[ - \dfrac{\pi }{2}\]
D. None of these

Answer 1

Hint: Here, an equation of definite integrals is given. First, rewrite the left-hand side by using the definite integral rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]. Then, simplify it by applying the trigonometric identities. After that, add this new integral with the original integral and simplify it. In the end, equate it with the right-hand side to get the required answer.

Formula Used: Definite integration rule: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]
If \[f\left( x \right) = f\left( {2a - x} \right)\] , then \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\]

Complete step by step solution: The given definite integral equation is \[\int\limits_0^\pi {xf\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx = k} \int\limits_0^{\dfrac{\pi }{2}} {f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \].

Let consider,
\[I = \int\limits_0^\pi {xf\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \] \[.....\left( 1 \right)\]
Apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the right-hand side.
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {{{\cos }^2}\left( {\pi - x} \right) + {{\tan }^4}\left( {\pi - x} \right)} \right)dx} \]
Now apply the trigonometric identities \[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\] and \[{\tan ^n}\left( {\pi - \theta } \right) = {\left( { - \tan \theta } \right)^n}\].
We get,
\[I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \] \[.....\left( 2 \right)\]

Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[2I = \int\limits_0^\pi {xf\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} + \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left( {x + \pi - x} \right)f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\pi f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \]
Here, \[f\left( x \right) = f\left( {2a - x} \right)\]. So, apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\].
\[ \Rightarrow 2I = 2\int\limits_0^{\dfrac{\pi }{2}} {\pi f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \]
Divide both sides by 2.
\[ \Rightarrow I = \pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \]

Now equate the above equation with the right-hand side of the given original equation.
\[\pi \int\limits_0^{\dfrac{\pi }{2}} {f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} = k\int\limits_0^{\dfrac{\pi }{2}} {f\left( {{{\cos }^2}x + {{\tan }^4}x} \right)dx} \]
Equating both sides, we get
\[k = \pi \]

Option ‘B’ is correct

Note: Sometimes students get confused about the exponent rule of the trigonometric ratios.
The formulas are as follows:
\[{\sin ^n}\left( {\pi - \theta } \right) = {\sin ^n}\theta \]
\[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\]
\[{\tan ^n}\left( {\pi - \theta } \right) = {\left( { - \tan \theta } \right)^n}\]