Question

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

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

Answer 1

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!

Answer 2

`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`