Question

If \[f\left( x \right)\] is an odd function of \[x\], then what is the value of \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {f\left( {\cos x} \right)} dx\]?
A. 0
B. \[\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\cos x} \right)} dx\]
C. \[2\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
D. \[\int\limits_0^\pi {f\left( {\cos x} \right)} dx\]

Answer 1

Hint: Here, a definite integral is given. First, check whether the given function \[f\left( {\cos x} \right)\] is an even function or an odd function. If the function is odd, then apply the integration property \[\int\limits_{ - a}^a {f\left( x \right)} dx = 0\]. If the function is even, then apply the integration property \[\int\limits_{ - a}^a {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] and then solve the integral by applying the integration property \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]. In the end, apply the trigonometric property \[\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x\] and simplify the integral to get the required answer.



Formula Used:\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function. Means, \[f\left( { - x} \right) = - f\left( x \right)\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function. Means, \[f\left( { - x} \right) = f\left( x \right)\]



Complete step by step solution:Given:
The definite integral is \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {f\left( {\cos x} \right)} dx\], where\[f\left( x \right)\] is an odd function of \[x\].
Let consider,
\[I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {f\left( {\cos x} \right)} dx\]
Now let’s check whether the function \[f\left( {\cos x} \right)\] is an odd function or an even function.
Let consider,
\[g\left( x \right) = f\left( {\cos x} \right)\]
Now calculate the value of \[g\left( { - x} \right)\]
\[g\left( { - x} \right) = f\left( {\cos \left( { - x} \right)} \right)\]
\[ \Rightarrow g\left( { - x} \right) = f\left( {\cos x} \right)\]
\[ \Rightarrow g\left( { - x} \right) = g\left( x \right)\]
 Thus, the function \[g\left( x \right) = f\left( {\cos x} \right)\] is an even function.

So, apply the integration rule \[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function.
We get,
\[I = 2\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\cos x} \right)} dx\]
Again, apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[I = 2\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\cos \left( {\dfrac{\pi }{2} - x} \right)} \right)} dx\]
Apply the trigonometric property \[\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x\].
\[I = 2\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\]
Thus, \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {f\left( {\cos x} \right)} dx = 2\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)} dx\].



Option ‘C’ is correct



Note: Sometimes students get confused about the exponent rule of the trigonometric ratios.
The formulas are as follows:
\[{\cos ^n}\left( {\dfrac{\pi }{2} - \theta } \right) = {\sin ^n}\theta \]
\[{\sin ^n}\left( {\dfrac{\pi }{2} - \theta } \right) = {\cos ^n}\theta \]