[SOLVED] Trig Integrals

**mollymcf2009** · February 5th 2009, 07:50 PM

Hey everyone!

Integrate.

$\int 59 cos^2 x \cdot sin (2x) dx$

Here is what I have done so far:

$59 \int (\frac{1}{2}(1 + cos 2x) \cdot (2 sinx cosx) dx$

Are those the correct identities to substitute in there?

Also need some help to go on to the next step of this.

Thanks y'all!

**Soroban** · February 5th 2009, 08:13 PM

Hello, mollymcf2009!

Integrate: . $59\int \cos^2\!x\sin2x\, dx$

Here is what I have done so far: . $59 \int \left(\frac{1 + \cos 2x}{2}\right)(2\sin x\cos x)\,dx$
Don't change both of them.

There are a few ways to approach this problem.
Here's one of them . . .

We have: . $59\int\cos^2\!x(2\sin x\cos x)\,dx \;=\;118\int\cos^3\!x(\sin x\,dx)$

Let: $u \,=\, \cos x \quad\Rightarrow\quad du \,=\, -\sin x\,dx \quad\Rightarrow\quad \sin x\,dx \,=\,-du$

Substitute: . $118\int u^3(-du) \;\;=\;\;-118\int u^3\,du \;\;=\;\;-118\,\frac{u^4}{4} + C \;\;=\;\;-\frac{59}{2}u^4 + C$

Back-substitute: . $-\frac{59}{2}\cos^4\!x + C$

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