
What is the value of the integral \[\int\limits_0^{2\pi } {{{\cos }^{99}}xdx} \]?
A. 1
B. \[ - 1\]
C. \[99\]
D. 0
Answer
161.7k+ views
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:
\[{\sin ^n}\left( {\pi - \theta } \right) = {\sin ^n}\theta \]
\[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\]
\[{\tan ^n}\left( {\pi - \theta } \right) = {\left( { - \tan \theta } \right)^n}\]
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:
\[{\sin ^n}\left( {\pi - \theta } \right) = {\sin ^n}\theta \]
\[{\cos ^n}\left( {\pi - \theta } \right) = {\left( { - \cos \theta } \right)^n}\]
\[{\tan ^n}\left( {\pi - \theta } \right) = {\left( { - \tan \theta } \right)^n}\]
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
