hi plizzzzzzzzz help... finding antiderivatives?

Question

how woud i find the antiderivative of 2Sin3x
and also the antiderivative of 1+2Cos2x

thanx!!!

Answer 1

Use the chain rule backwards in both cases.
Antiderivative of 2 sin 3x = -2/3 cos 3x + C.
For the second one, do each part separately:
Antiderivative of 1 + 2cos 2x = x + sin 2x + C.

Answer 2

Here's the first:

use S for the the antiderivative sign

∫ 2sin3x dx

we only have a rule for ∫ sin x dx

with no "3x" in the argument for sin. so we just use substitution to get:

I'll use Ó¨ for theta.

let Ó¨ = 3x, then dÓ¨ = 3 dx, and solving for dx, dx = dÓ¨/3

so plug this back in to get:

∫ 2 sin Ө dӨ/3

taking out the constants gives:

2/3 ∫ sin Ө dӨ

now the antiderivative of sin Ó¨ dÓ¨ is just -cosÓ¨ (don't for get the minus sign), so we get:

-2/3 cos Ó¨

but we don't want Ó¨ we want x so just substitute again to get:

-2/3 cos 3x

taking the derivative of this will give you 2sin3x, so it's correct.

Here's the second:

∫ 1+2cos2x dx

∫ 1 dx + S 2cos2x dx

x + 2 ∫ cos 2x dx

now do 2x = Ó¨, then take derivative of both sides to get:

2 dx = dÓ¨

dx = dÓ¨/2

x + 2 ∫ cos Ө dӨ /2

x + 2/2 ∫ cos Ө dӨ

x + ∫ cos Ө dӨ

x + sin Ó¨

x + sin 2x

Answer 3

Well, you need to know the basics here.

antiderivative F of cos(nx) = n*sin(nx) + k
antiderivative F of sin(nx) = -n*cos(nx) + k

where k is any constant number you need to write for completeness....

Using this information, you may now do the above.