Question

What is the value of the integral \[\int\limits_0^{2\pi } {{{\cos }^{99}}xdx} \]?
A. 1
B. \[ - 1\]
C. \[99\]
D. 0

Answer 1

Hint: Here, a definite integral is given. First, simplify the integral by applying the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]. Then, apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] and simplify the integral. After that, add both integrals and solve it to get the required answer.



Formula Used:Integration rules:
\[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]



Complete step by step solution:The given definite integral is \[\int\limits_0^{2\pi } {{{\cos }^{99}}xdx} \].
Let consider,
\[I = \int\limits_0^{2\pi } {{{\cos }^{99}}xdx} \]
Simplify the integral by applying the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
 \[I = 2\int\limits_0^\pi {{{\cos }^{99}}xdx} \] \[.....\left( 1 \right)\]
Now apply the integration rule\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
We get,
\[I = 2\int\limits_0^\pi {{{\cos }^{99}}\left( {\pi - x} \right)dx} \]
Apply the trigonometric formula \[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\].
\[ \Rightarrow I = 2\int\limits_0^\pi { - {{\cos }^{99}}xdx} \]
\[ \Rightarrow I = - 2\int\limits_0^\pi {{{\cos }^{99}}xdx} \] \[.....\left( 2 \right)\]

Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = 2\int\limits_0^\pi {{{\cos }^{99}}xdx} - 2\int\limits_0^\pi {{{\cos }^{99}}xdx} \]
\[ \Rightarrow 2I = 0\]
\[ \Rightarrow I = 0\]
Therefore, \[\int\limits_0^{2\pi } {{{\cos }^{99}}xdx} = 0\].



Option ‘D’ is correct


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