yet ANOTHER integral that i can't grasp...?

Question

int (e^-theta)*cos(2theta)

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Answer 1

Let u = theta
integral e^(au) cos nu du = (e^au(acos nu + n sin nu))/(a^2+n^2) + C
In your case a = -1, n = 2, so we get:
(e^-u(-cos 2u +2 sin 2u))/5 + C

Answer 2

I'm going to call theta 't'
e^-t * cos (2t)

***Integration by parts
u = e^-t dv = cos(2t)
du = -e^-t dt v = (sin(2t) ) / 2

int(e^-t * cos (2t)) = [e^-t * sin(2t)] / 2 + 1/2 * int(sin(2t) * e^-t)

***Integration by parts once more
u = e^-t dv = sin(2t)
du = -e^-t d t v = -(cos(2t) ) / 2

int(e^-t * cos (2t)) = [e^-t * sin(2t)] / 2 + 1/2 * { -[e^-t * cos(2t)] / 2 - 1/2 int (cos(2t) * e^-t)} + C

***Add [1/2 int (cos(2t) * e^-t)] to both sides
3/2 * int(e^-t * cos (2t)) = [e^-t * sin(2t)] / 2 + 1/2 * { -[e^-t * cos(2t)] / 2} + C

***Divide both sides by 3/2
int(e^-t * cos (2t)) = 2/3 * { [e^-t * sin(2t)] / 2 + 1/2 * { -[e^-t * cos(2t)] / 2} + C}

Then simplify. Hope this helps!

Answer 3

Rather than solve it for you, let me just tell you that a similar tactic is used to integrate sec³(x). It uses integration by parts, and uses a "circular" method of obtaining the answer. I'll show you the example of the similar question, and perhaps you can work it out on your own.

Example:

∫sec³(x) dx

First, split sec³(x) into sec(x) and sec^2(x).

∫ sec(x) sec²(x) dx

Use integration by parts.

Let u = sec(x). dv = sec²(x) dx
du = sec(x)tan(x) dx. v = tan(x)

sec(x)tan(x) - ∫sec(x)tan²(x)dx

Use the identity tan²(x) = sec²(x) - 1

sec(x)tan(x) - ∫sec(x)[sec²(x) - 1] dx

Distribute the sec(x).

sec(x)tan(x) - ∫(sec³(x) - sec(x)) dx

Separate into two integrals,

sec(x)tan(x) - [∫sec³(x)dx - ∫sec(x)dx]

One of the integrals worth remembering is ∫sec(x)dx = ln|sec(x) + tan(x)|. Substituting that, we get

sec(x)tan(x) - [∫sec³(x)dx - ln|sec(x) + tan(x)|]

Distribute the minus sign.

sec(x)tan(x) - ∫sec³(x) dx + ln|sec(x) + tan(x)|

At this point, it would appear that we've gone in a circle; that isn't the case, because if we look at the whole equation again,

∫sec³(x)dx = sec(x)tan(x) - ∫sec³(x)dx + ln|sec(x) + tan(x)|

What we're going to do at this point is *ADD* ∫sec³(x)dx to both sides as if it were a variable. This eliminates the -∫sec³(x)dx on the right hand side, and makes 2 of them on the left hand side.

2∫sec³(x)dx = sec(x)tan(x) + ln|sec(x) + tan(x)|

Now, we multiply both sides by (1/2), to get rid of the 2 on the left hand side.

∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)|

And, don't forget to add a constant. Our final answer is then:

∫sec³(x)dx = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C