Calculate int_0^(2pi) cos^2x .dx?

A) Zero
B) 2
C) 1/2
D) pi
The answer is D for your reference

2 Answers
Sep 5, 2017

Yes, the correct answer is D, or pi.

Explanation:

Use the power reduction formula

cos^2x= (1 + cos2x)/2.

So we have:

I = int_0^(2pi) (1 + cos2x)/2

I = 1/2int_0^(2pi) 1 + cos(2x)dx

I = 1/2int_0^(2pi) 1 dx + 1/2int_0^(2pi) cos(2x)dx

If we let u = 2x in the second integral, then 1/2du = dx. The integral of cosu is sinu and reverse your substitution to get sin(2x).

I = 1/2[x]_0^(2pi) + 1/2(1/2)[sin(2x)]_0^(2pi)

The integral equals

I = 1/2(2pi) + 1/4(sin(4pi) - sin(0))

I = pi - 1/4(0 - 0)

I = pi

Answer D, as required.

Hopefully this helps!

Sep 6, 2017

int_0^(2pi)cos^2(x)dx

We can rewrite this:

int_0^(2pi)cos^2(x)dx=int_0^(2pi)(cos^2(x)-sin^2(x)+sin^2(x))dx

Note that cos(2x)=cos^2(x)-sin^2(x):

int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)sin^2(x)dx

Rewrite sin^2(x) using sin^2(x)+cos^2(x)=1:

int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)(1-cos^2(x))dx

int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)dx-int_0^(2pi)cos^2(x)dx

Add int_0^(2pi)cos^2(x)dx to both sides of the equation:

2int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)dx

Evaluate these integrals:

2int_0^(2pi)cos^2(x)dx=|1/2sin(2x)|_0^(2pi) +x|_0^(2pi)

2int_0^(2pi)cos^2(x)dx=1/2(sin(4pi)-sin(0))+(2pi-0)

2int_0^(2pi)cos^2(x)dx=2pi

Dividing by 2:

int_0^(2pi)cos^2(x)dx=pi