Confused on how to do this (Integration C4)

Alright so I'm confused on how to integrate this with recognition / applying chain rule in reverse.

I know it can be solved with other methods eg: IBS, but would like to know how to do it, the other way.

Reply 1

B_9710

18

Consider the derivative of

\sin(x^2)

.

Reply 2

Zacken

22

Original post by SaadKaleem

Alright so I'm confused on how to integrate this with recognition / applying chain rule in reverse.

I know it can be solved with other methods eg: IBS, but would like to know how to do it, the other way.

The important thing to note here is that integration by substitution is exactly the same thing as recognition/applying the chain rule in reverse, just a slightly different way of looking at it.

So here, you can see you have an integrand of the form

\frac{d}{dx}(x^2) \cos (x^2) \, \mathrm{d}x

. So you can immediately see that it'll integrate to something like

\sin(x^2)

(up to a constant factor blah blah).

So what you do is

\frac{d}{dx}(\sin (x^2)) = 2x \cos (x^2)

Your integrand is

\frac{1}{3}x\cos (x^2)

so you want to turn the 2 into a 1/3.

The way to do this is:

\frac{1}{3 \times 2}\frac{d}{dx}(\sin(x^2)) = \frac{1}{3}x\cos (x^2)

So integrating

\frac{1}{3}x\cos (x^2)

gives you

\frac{1}{6}\sin(x^2)

by recognition.

With enough practice, you'll be able to skip the step of writing down

\frac{d}{dx}(sin (x^2)) = 2x\cos (x^2)

and be able to jump straight to the answer, I will stress that this comes with practice, so for now - stick with writing the derivative.

Reply 3

edothero

15

Original post by SaadKaleem

Alright so I'm confused on how to integrate this with recognition / applying chain rule in reverse.

I know it can be solved with other methods eg: IBS, but would like to know how to do it, the other way.

Alright so I usually have a hard time explaining integration by recognition but here goes.

First of all, take some sort of a "guess". Smart guess of course.

With this integral I would say

y=sin(x^{2})

As the integral of cosx is sinx
This line is basically saying my guess is that this will integrate to

sin(x^{2})

Now, once you've made this guess, you differentiate it to see if you get what you're trying to integrate.

y=sin(x^{2})

\dfrac{dy}{dx} = 2xcos(x^{2})

You can see this is very close to the main integral, but its 6 times too big.

So from that you will know that the main question integrates to

\dfrac{1}{6}sin(x^{2})

I hate explaining integration by recognition, I find it hard to explain :laugh:

Though Zacken has beat me to it anyway :biggrin:

Reply 4

SaadKaleem

OP

8

Original post by B_9710

-----.

Original post by Zacken

-----

Original post by edothero

-----

Wow thank you all for the fast responses, yeah that was a bit dumb moment for me, Should've considered sin(x^2)... lol :frown:

Anyways, I've understood it. :smile:

Reply 5

Zacken

22

Original post by SaadKaleem

Wow thank you all for the fast responses, yeah that was a bit dumb moment for me, Should've considered sin(x^2)... lol :frown:

Anyways, I've understood it. :smile:

Hmm...

2x\cos(x^2) \neq \frac{1}{6} \cdot 2x \cos(x^2)

.

Reply 6

SaadKaleem

OP

8

Original post by Zacken

Hmm...

2x\cos(x^2) \neq \frac{1}{6} \cdot 2x \cos(x^2)

.

Damn, I've been making that mistake, I guess.

But can't it be used as a check that multiplying by 1/6 would equal the dy/dx value to the original integrand? So, we'd now also need to multiply by 1/6 to the value we've considered?

Reply 7

Zacken

22

Original post by SaadKaleem

Damn, I've been making that mistake, I guess.

But can't it be used as a check that multiplying by 1/6 would equal the dy/dx value to the original integrand? So, we'd now also need to multiply by 1/6 to the value we've considered?