Question

What is `int cos^2(3u)du`?

Answer 1

The answer is `=1/2u+1/12sin(6u)+C`

Explanation:

Reminder

`cos(2x)=2cos^2(x)-1`

`cos^2(x)=1/2(1+cos(2x))`

`cos^2(3x)=1/2(1+cos(6x))`

Therefore,

`intcos^2(3u)=1/2int(1+cos(6u))du`

`=1/2(u+1/6sin(6u))`

`=1/2u+1/12sin(6u)+C`

Answer 2

We seek:

`int \ cos^2(3u) \ du = (6u + sin 6u)/12 + C`

Explanation:

We seek:

`I = int \ cos^2(3u) \ du`

We can use the trigonometric identity:

`cos 2x -= cos^2x - sin^2 x`

From which we get:

`cos 2x -= cos^2x - (1-cos^2x) <=> cos^2x -= (1+cos 2x)/2`

So, we can write the integral as:

`I = int \ (1+cos 6u)/2 \ du`

`\ \ = 1/2 \ int \ 1+cos 6u \ du`

`\ \ = 1/2 \ {u + (sin 6u)/6} + C`

`\ \ = 1/2 \ {(6u + sin 6u)/6} + C`

`\ \ = (6u + sin 6u)/12 + C`