/sci/ - Science & Math

btw i'm a girl

>>4784425
<span class="math"> \int \cos^2 (\alpha) d\alpha [/spoiler]

just so it's complete and I don't look even sillier.

Bumping.

>>4784430
Also, I'll just be using a tripcode because some people think that trolling on 4chan has ever been funny.

You just have to do a lot of them and develop an intuition.

>>4784440
Yeah, that's what thought. I'll just learn all of the trigonometric derivations first, I suppose..

you could also think of integrals as areas under the curve you are integrating and how this area changes...

>>4784430
Completely irrelevant.
btw I'm a boy.

See how dumb it sounds?

Tip: use a computer. That's what people in the field do.

>>4784446
Yeah so?

>>4784447
It seems like you didn't read my third post - wasn't me.

Normally I'd use a computer, but in our exams (it's fairly early in the curriculum), we aren't allowed to because they want us to know the basics first.

ok so its accutally pretty easy just algebra intensive
first remember
e^{ix} = \cos x + i \sin x and thus
e^{-ix} = \cos (-x) + i \sin(-x) ==> \cos (x) - i \sin (x)

solve for cos(x) yeilds

\cos (x) = \frac{e^{ix}+e^{-ix}}{2}

sub into original
\int \cos^2 (x) dx ==> \int \left ( \frac{e^{ix}+e^{-ix}}{2} \right )^2

\int \frac{1}{4} \left ( e^{2ix} + 2 + e^{-2ix} \right )

\frac{1}{4} \left ( \frac{1}{2i} e^{2ix} +2x - \frac{1}{2i} e^{-2ix} \right )

sub back in definitions

\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x \right ) +2x - \frac{1}{2i} \left ( \cos -2x + i \sin -2x \right ) \right )


\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x \right ) +2x - \frac{1}{2i} \left ( \cos 2x - i \sin 2x \right ) \right )


\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x \right ) +2x + \frac{1}{2i} \left (- \cos 2x + i \sin 2x \right ) \right )

reduces to

\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x - \cos 2x +i \sin 2x \right ) \right )

and

\frac{1}{4} \left ( \sin 2x + 2x \right )


this method is rightly called complexification

>>4784425

there is no one rule fits all for integrals.

some are unsolvable ie integral of e^(x^2) dx

there are some special cases
simple --> use reverse power rule, trig identities or other identities, tables etc (ie (cos(x))^2, x^3, e^x, sin(x), cosh (x))

by parts ---> for products and other special cases ( ln x, cos(x)e^(x), x^2sin(x))

rational function (denominator factorable) use PFD

denominator (quadratic in denominator, un factorable) - complete the square

trig substitution ---> useful for some quadratic rational functions

substitution--> something in the integral's derivative appears again.

there are all trivial methods though

<span class="math"> e^{ix} = \cos x + i \sin x
?[/spoiler]

Experience will give you the right intuition. There are only so many tricks out there, and only certain types of functions that you can find a closed-form antiderivative for in terms of elementary functions. For example you will learn to recognize things like
<span class="math">exp(-x^2)[/spoiler]
which don't have nice antiderivatives while polynomials in sin and cos can be solved with trig tricks. In your case
<span class="math">cos^2(x) = \frac{1}{2} cos(2x) + \frac{1}{2} [/spoiler]

plus a CONSTANT!

sorry for the double post hopefully this time i accidentally got the correct math script

<span class="math"> e^{ix} = \cos x + i \sin x and thus
<span class="math"> e^{-ix} = \cos (-x) + i \sin(-x) ==> \cos (x) - i \sin (x)

solve for cos(x) yeilds

\cos (x) = \frac{e^{ix}+e^{-ix}}{2}

sub into original
\int \cos^2 (x) dx ==> \int \left ( \frac{e^{ix}+e^{-ix}}{2} \right )^2

\int \frac{1}{4} \left ( e^{2ix} + 2 + e^{-2ix} \right )

\frac{1}{4} \left ( \frac{1}{2i} e^{2ix} +2x - \frac{1}{2i} e^{-2ix} \right )

sub back in definitions

\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x \right ) +2x - \frac{1}{2i} \left ( \cos -2x + i \sin -2x \right ) \right )


\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x \right ) +2x - \frac{1}{2i} \left ( \cos 2x - i \sin 2x \right ) \right )


\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x \right ) +2x + \frac{1}{2i} \left (- \cos 2x + i \sin 2x \right ) \right )

reduces to

\frac{1}{4} \left ( \frac{1}{2i} \left ( \cos 2x + i \sin 2x - \cos 2x +i \sin 2x \right ) \right )

and

\frac{1}{4} \left ( \sin 2x + 2x \right )


this method is rightly called complexification[/spoiler][/spoiler]

>>4784806
zero is a constant

Thanks for the useful posts!

>>4784425
I actually had this problem in my book.
By using trig identities you'll see that.
cos(x)^2 = (1 + cos(2x)) / 2

This is hopefully easier to integrate.

>>4784430
Am I supposed to want to help you more just because you state that you are a girl?

I'm stubborn and the easiest way to get me to not do something is to try and manipulate me like that.
That being said, I would be worlds farther away from ever helping you JUST because you made that post.
Fuck off.

>>4784869
Mah nigga.

>>4784869
that wasn't OP, that was a troll