Question

If \[P = \int\limits_0^{3\pi } {f\left( {{{\cos }^2}x} \right)dx} \] and \[Q = \int\limits_0^\pi {f\left( {{{\cos }^2}x} \right)dx} \], then which of the following equation is true?
A. \[P - Q = 0\]
B. \[P - 2Q = 0\]
C. \[P - 3Q = 0\]
D. \[P - 5Q = 0\]

Answer 1

Hint: Here, two definite integrals are given. First, simplify the first given integral by checking the integral formula \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( {2a - x} \right) = f\left( x \right)\]. Then, verify the relation and calculate the required answer.



Formula Used:Integration rule: \[\int\limits_0^{n\pi } {f\left( x \right)} dx = n\int\limits_0^\pi {f\left( x \right)} dx\] if \[f\left( {x + n\pi } \right) = f\left( x \right)\]
Trigonometric identity: \[{\cos ^n}\left( {x + \pi } \right) = {\left( { - \cos x} \right)^n}\]



Complete step by step solution:The given definite integrals are:
\[P = \int\limits_0^{3\pi } {f\left( {{{\cos }^2}x} \right)dx} \] \[.....\left( 1 \right)\]
\[Q = \int\limits_0^\pi {f\left( {{{\cos }^2}x} \right)dx} \] \[.....\left( 2 \right)\]

Let consider,
\[g\left( x \right) = f\left( {{{\cos }^2}x} \right)\]
Substitute \[x = x + \pi \] in the above equation.
We get,
\[g\left( {x + \pi } \right) = f\left( {{{\cos }^2}\left( {x + \pi } \right)} \right)\]
Apply the trigonometric formula \[{\cos ^n}\left( {x + \pi } \right) = {\left( { - \cos x} \right)^n}\].
\[g\left( {x + \pi } \right) = f\left( {{{\left( {\cos x} \right)}^2}} \right)\]
\[ \Rightarrow g\left( {x + \pi } \right) = f\left( {{{\cos }^2}x} \right)\]
\[ \Rightarrow g\left( {x + \pi } \right) = g\left( x \right)\]
Here, \[g\left( x \right)\] is a periodic function with period \[\pi \].
The integration formula for a periodic function is,
\[\int\limits_0^{n\pi } {f\left( x \right)} dx = n\int\limits_0^\pi {f\left( x \right)} dx\]
So, for the equation \[\left( 1 \right)\] we get
 \[P = \int\limits_0^{3\pi } {f\left( {{{\cos }^2}x} \right)dx} \]
\[ \Rightarrow P = 3\int\limits_0^\pi {f\left( {{{\cos }^2}x} \right)dx} \]
Substitute equation \[\left( 2 \right)\] in the above equation.
 \[ \Rightarrow P = 3Q\]
\[ \Rightarrow P - 3Q = 0\]



Option ‘C’ is correct



Note: Students often get confused with trigonometric identities.
Remember the following trigonometric identities:
\[{\sin ^n}\left( {x + \pi } \right) = {\left( { - \sin x} \right)^n}\]
\[{\cos ^n}\left( {x + \pi } \right) = {\left( { - \cos x} \right)^n}\]
\[{\tan ^n}\left( {x + \pi } \right) = {\tan ^n}x\]