why do we have to switch sin into cos

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[olive nimbus]

Can’t we just make u equal to sin(x) and use u-substitution?

[swift socket]

Well they linearised x—>sin^2(x) to integrate more easily

You can integrate x—>cos(2x) easily

[olive nimbus]

we get different answers tho

[swift socket]

What do you mean ?

[olive nimbus]

sin^2 (x)= sin^3(x)/3 if we use U substitution and if we plug in pi and 0 for the intergral we get 0

but if we solve the other problem we end up with pi/2

[swift socket]

? Idk where you got that first equality but regardless substitution doesn’t work here because u—>arcsin(u) has values in [-pi/2, pi/2] you can’t use the substitution theorem here

By definition if [a,b] is a closed interval and f a continuous function on that closed interval then, if $\varphi$ is a class C1 bijective function from $[\alpha, \beta]$ such that $\varphi(\alpha)=a$ and $\varphi(\beta)=b$ then $\int_{a}^{b} f(t)dt= \int_{\alpha}^{\beta} \varphi^{‘}(u) f(\varphi(u))du$

[knotty moat]

Rotor

[swift socket]

Here substitution doesn’t work because setting x=arcsin(u), u—>arcsin(u) isn’t even surjective

[swift socket]

To [a,b]

[sharp lake]

That's not remotely true. You forgot to replace dx with du.

[olive nimbus]

Oh thats what i probarly forgot, so I don't have to do what he did and switch sin^2x into (1-cos2x)/2

[sharp lake]

I mean, you don't have to do that specifically, but u sub still doesn't work. Also I don't know why you have such a problem with this conversion.

[olive nimbus]

whats the problem?

[sharp lake]

...I don't know. You're the one who has it, you tell me.

[olive nimbus]

I don't have a problem....

And why doesn't u sub not work?

[swift socket]

Yes just use this

[sharp lake]

...if you don't have a problem with it, why do you want so badly not to use it?

[olive nimbus]

because I like to keep it simple and automated

to be able to use U-sub constantly is nice

[sharp lake]

This is simple.

Also, that's just not how integration works, sorry.

[swift socket]

I gave an explanation (albeit not clear) but just know that it doesn’t work here

[olive nimbus]

converting it into cos is not simple, although the process is simple just having that idea isn't so straight foward

[swift socket]

Well it’s just simple trig identities

[olive nimbus]

@swift socket I don't understand mathmatical languange to be honest

Would you tell me in leymans term

why it doesnt work?

[sharp lake]

You're literally saying you don't want to have to think.

[sharp lake]

Because when you actually remember to do the dx/du substitution, the function you get isn't any easier to integrate than the function you started with.

And also because of what Rotor is saying.

[olive nimbus]

Oh so is it because that

we get dx=du/cos(x)

[sharp lake]

And also because of what Rotor is saying.

[olive nimbus]

I don't understand what Rotor is saying tbh

my mathmatical termnology isn't the best

thats why i asked him to dumb it down

[sharp lake]

I know you don't, and I can't figure out how to explain it to you, so I'm giving you a second explanation that you can understand so you'll give up on the idea that doesn't work.

[olive nimbus]

I understand.

okay

thanks.

[sharp lake]

Actually, here's a way you might understand. What's sin(0)?

[olive nimbus]

0

[sharp lake]

What's sin(pi)?

[olive nimbus]

0

[sharp lake]

So if u = sin(x), you'd be integrating from u = 0 to... u = 0.

[olive nimbus]

and you cant intergrate a point, so therefore we need to switch sin into cos

correct?

[sharp lake]

Well, the integral of a point is zero, but this integral obviously isn't.

[olive nimbus]

Oh and wouldn't be zero as if we look at the sin graph we can obviously see an area underneath the curve

[sharp lake]

In fact, it's impossible for any integral of this function to ever be zero as it's always nonnegative.

[olive nimbus]

thanks man!

[sharp lake]

Do you need the actual conversion explained to you?

[olive nimbus]

nope I compely understand it's just trig identities

completely*

[sharp lake]

Okay, but do you understand how it's derived? Like, the actual sequence of steps to derive it?

[olive nimbus]

Are you asking how sin^2(x)=(1-cos(2x))/2 is derived?

if you are than no

[swift socket]

Do you know what cos(a+b) is ?

With respect to cos(a), cos(b) sin(a) and sin(b)?

From this you can deduce cos(2x) with respect to cos(x) and sin(x) and using the formula cos^2(x)+sin^2(x)=1 you have the result

[sharp lake]

Okay, so. You know cos^2(x) + sin^2(x) = 1, right?

[olive nimbus]

yes

[sharp lake]

Do you know the double angle formula for cosine?

[olive nimbus]

yes

2cos^2(x)-1

[sharp lake]

What's another form for it?

[olive nimbus]

cos^2x-sin^2x

[sharp lake]

Right. So we have: cos^2(x) + sin^2(x) = 1 cos^2(x) - sin^2(x) = cos(2x)We want an identity for sin^2(x), so we subtract the second equation from the first one, and what do we get?

Actually, a simpler way is just to note cos(2x) = 1 - 2sin^2(x) and go from there.

[olive nimbus]

(cos(2x)-1)/2=sin^2(x)

[sharp lake]

Incorrect.

You ought to have divided by -2, not 2.

[olive nimbus]

Oh okay

(Cos(2x)-1)/-2=sin^2(x)

[sharp lake]

...so then what happens?

[olive nimbus]

+close