Evaluating the Integral:?

Question

∫cos^3(3x+4)dx

I'm a bit confused with this question, any help would be much appreciated.

Thanks,
Shaun

Answer 1

∫[cos^3(3x+4)dx]

Let's solve this using substitution twice.

Let u = 3x + 4. Then
du = 3 dx, so
(1/3) du = dx

This makes our integral

∫[cos^3(u) (1/3) du]

Pulling the constant (1/3) out of the integral,

(1/3) ∫ cos^3(u) du

To solve this integral, break off a cos(u).

(1/3) ∫ [cos^2(u) cos(u) du]

Now, use the identity cos^2(u) = 1 - sin^2(u).

(1/3) ∫ [(1 - sin^2(u)) cos(u) du]

Here's where we do substitution again.
Let v = sin(u). Then
dv = cos(u) du {Note: look at how the tail end of our integral is
cos(u) du. This will become dv after the substitution.}

So now we have

(1/3) ∫ (1 - v^2) dv

And now, this integral is trivial. Using the reverse power rule, we get

(1/3) [v - (1/3)v^3] + C

Distributing (1/3) over the brackets,

(1/3)v - (1/9)v^3 + C

Going backwards from our substitution, since we let v = sin(u), then

(1/3)sin(u) - (1/9) sin^3(u) + C

And going backwards further, since u = 3x + 4, then we have

(1/3) sin(3x + 4) - (1/9) sin^3(3x + 4) + C

Answer 2

Try substitution. 3x+4=u. du=3dx. dx=du/3.

∫cos^3(3x+4)dx=

1/3∫cos^3(u)du

Then I'm pretty sure you can use integration by parts to solve this problem: cos^3(u)=w, du=dv. You should probably try doing this part by yourself, it's your homework :)

Answer 3

♠ y(x)= cos^3(3x+4), assume t=3x+4 and dx= dt/3;
(cost)^3=cost*(cost)^2 = 0.5cost*(cos2t+1) =
={see! cosA*cosB=0.5(cos(A+B) +cos(A-B)} =
= 0.5cost +0.25(cos3t +cost) =0.75cost +0.25cos3t;
♣ I=∫y(t)dt/3 = 0.75∫cost·dt/3 +0.25∫cos3t·dt/3 =
= 0.25sint +(0.25/9)·sin3t =0.25(sin(3x+4) +(1/9)·sin(9x+12)) +C;
Puggy’s also correct!