# Integral! Cal. 2!

$$\int 96cos^4(6x) * dx$$

ok first i take out the 96 cause it's constant!

$$96 \int cos^4(6x) * dx$$

$$96 \int (cos^2(6x)^2) * dx$$

ok now with that setup, i can know use the half-angel formula!

$$96 \int (\frac{1+cos(12x)}{2})^2$$

squared the problem...

$$24 \int (1+ 2cos(12x) + (cos12x)^2$$

now to use the half-angle idents agian..

$$24 \int 1 +2cos(12x) + 1/2(1+cos24x)$$

can someone tell me if im doing this correctly before i integral the problem?

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## Answers and Replies

Pyrrhus
Homework Helper
Yes use Half Angle Identities.

$$cos^2(6x) = \frac{1+cos(12x)}{2}$$

$$\int 96cos^4(6x) dx$$

$$96\int \frac{1+cos(12x)}{2} \frac{1+cos(12x)}{2}dx$$

$$96\int \frac{(1+cos(12x))^2}{4} dx$$

$$24[\int 1 dx + \int 2cos(12x) dx + \int cos^2(12x) dx]$$

You are doing fine...

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Cyclovenom said:
Yes use Half Angle Identities.

$$cos^2(6x) = \frac{1+cos(12x)}{2}$$

$$\int 96cos^4(6x) dx$$

$$96\int \frac{1+cos(12x)}{2} \frac{1+cos(12x)}{2}dx$$

$$96\int \frac{(1+cos(12x))^2}{4} dx$$

Expand and etc... keep going

Edit:Sleep got to me...

can i take out the 1/4? so 1/4*96 = 24

Pyrrhus
Homework Helper
Whatupdoc said:
can i take out the 1/4? so 1/4*96 = 24
yes you can

thank you for checking my answer, ok i will cont. where i left off....

$$24 \int 1 +2cos(12x) + 1/2(1+cos24x)$$

$$24 \int 1 +2cos(12x) + 1/2+\frac{1}{2}cos(24x)$$

$$24 \int1dx + 48 \int cos(12x) + 12 \int dx + 12 \int cos(24x)$$

adding alot of integral signs seems to look messy, but i think you can tell what im doing. ok time to find the anti-dervs.

$$24x + \frac{1}{12}(48)sin(12x) + 12 + 12(\frac{1}{24})sin(24x)$$

k time to solve...
$$24x+4sin(12x) + 12 + \frac{1}{2}sin(24x)$$

but it's the wrong answer

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Whatupdoc said:
expand this alittle more
$$1/2(1+cos24x) = 1/2+cos12x$$ right? <-- now sure about this part

This bit is wrong. 1/2(1 + cos 24x) = 1/2 + (1/2)cos 24x. (1/2)cos x isn't the same thing as cos (x/2).

Nylex said:
This bit is wrong. 1/2(1 + cos 24x) = 1/2 + (1/2)cos 24x. (1/2)cos x isn't the same thing as cos (x/2).

ah i see, dont know what i was thinking. i edited my previous post and fixed the problem, but the answer is still wrong

Pyrrhus
Homework Helper
$$24[\int 1 dx + \int 2cos(12x) dx + \int cos^2(12x) dx]$$

$$24[\int 1 dx + \int 2cos(12x) dx + \int \frac{1+cos(24x)}{2} dx]$$

$$24\int 1 dx + 24\int 2cos(12x) dx + 24\int \frac{1+cos(24x)}{2} dx$$

$$24\int 1 dx + 48\int cos(12x) dx + 12\int 1 dx+ 12\int cos(24x) dx$$

$$24x + 4sin(12x) + 12x+ \frac{sin(24x)}{2} + C$$

$$36x + 4sin(12x)+ \frac{sin(24x)}{2} + C$$

Check my work...

Cyclovenom said:
$$24[\int 1 dx + \int 2cos(12x) dx + \int \frac{1+cos(24x)}{2} dx]$$

$$24\int 1 dx + 24\int 2cos(12x) dx + 24\int \frac{1+cos(24x)}{2} dx$$
What in the world are you doing here!? Where did those "+ 24"s come from?

Nevermind, I was reading it wrong (since it looks cluttered).[/edit]

OK. I checked your last calculation. Everything is correct now.

Pyrrhus
Homework Helper
Thanks but i meant that to whatuptdoc

ah i see what i did wrong, thank you