Integration by Substiution

Original post by Illidan2

Where's your working?

Reply 2

Sorry, it's mid-upload. You should see it in a minute or so. If you need any further clarity on what i've written, please let me know :smile:

Edit: It should now be visible on the imgur link

(edited 6 years ago)

Reply 3

Original post by Illidan2

Sorry, it's mid-upload. You should see it in a minute or so. If you need any further clarity on what i've written, please let me know :smile:

Edit: It should now be visible on the imgur link

u^2=e^x-2

does not mean that

2\dfrac{du}{dx} = e^x

. Instead, you should differentiate implicitly on the LHS. If this is C3, then just root both sides before differentiating.

Reply 4

the bear

you should get

2u*du/dx = e^x

Reply 5

I'm self-taught on the new spec, on the penultimate chapter of the year two pure maths book- I imagine this is the equivalent of C4 Integration. I did not realise it didn't mean that, I saw something like this in a solution earlier today and tried using it here, thinking it would make it easier for myself- I wish i'd square-rooted it like you said, as I would have if I hadn't tried that.

Reply 6

Original post by the bear

you should get

2u*du/dx = e^x

Yes, that's it! I meant 2u. >.>

Reply 7

Original post by Illidan2

Also this is wrong. You cannot (a) magically take the reciprocal of the inside, and (b) apply the log integral like that.

(edited 6 years ago)

Reply 8

Original post by the bear

you should get

2u*du/dx = e^x

Aside from this, if I were to redo the entire thing in the way i've done it so far, only adjusting everything to accomodate for 2u*du/dx = e^xinstead of 2du/dx=e^x, will I still be able to reach the answer? I did some implicit differentiation the other day but i've only done it a few times and I can't remember exactly how it's done.

Reply 9

the bear

Original post by Illidan2

it comes out really quickly if you stick with your original method...

Reply 10

Original post by the bear

it comes out really quickly if you stick with your original method...

When you say my original method, what are you referring to?

Reply 11

Original post by RDKGames

Also this is wrong. You cannot (a) magically take the reciprocal of the inside, and (b) apply the log integral like that.

Yeah. The reciprocal thing I can see now. As for the log application, if the integral of 1/x dx= ln |x| +c, can I not use it like that as long as I have the integral of 1 divided by a term?

(edited 6 years ago)

Reply 12

Original post by Illidan2

Yeah. The reciprocal thing I can see now. As for the log application, if 1/x dx= ln |x| +c, can I not use it like that as long as I have 1 divided by a term?

That's just one application of the main rule.

The main rule you need to know and be able to use is

\displaystyle \int \dfrac{f'(x)}{f(x)} .dx = \ln |f(x)|+c

. You do not have this form in your incorrect working anyway, so you cannot use the rule.

Reply 13

the bear

Original post by Illidan2

When you say my original method, what are you referring to?

the substitution

Reply 14

I thought my entire method WAS the substitution. I thought that once i'd got it into the correct form with u, I could substitute it back for the x-values and then compare with the form mentioned in the question to find a, b, c and d. I've confused myself now, it seems.

Reply 15

Original post by RDKGames

That's just one application of the main rule.

The main rule you need to know and be able to use is

\displaystyle \int \dfrac{f'(x)}{f(x)} .dx = \ln |f(x)|+c

. You do not have this form in your incorrect working anyway, so you cannot use the rule.

Yeah, i've seen the rule before. I've yet to commit it to memory, though. I'm now starting again to see if I can get anywhere with this.

Reply 16

Okay. So i've written out a few of the substitutions from the start again. Now, what should my approach be from here?

https://imgur.com/a/N2AOn

Edit: I know the differentiation is wrong. Via the chain rule, I should have a multiplier of 4 in there, too, right?

(edited 6 years ago)

Reply 17

Original post by Illidan2

\displaystyle \int_{\ln_3}^{\ln 4} \dfrac{e^{4x}}{e^x-2}.dx

Let

u^2 = e^x-2 \implies 2udu=e^x dx \implies dx = \dfrac{2u}{e^x}.du

So we have

\displaystyle \int_1^{\sqrt{2}} \dfrac{e^{4x}}{u^2} \cdot \frac{2u}{e^x} .du = 2\int_1^{\sqrt{2}} \frac{(u^2+2)^3}{u}.du = 2 \int_1^{\sqrt{2}} \frac{u^6+6u^4 +12 u^2 +8}{u}.du

Reply 18

Original post by RDKGames

\displaystyle \int_{\ln_3}^{\ln 4} \dfrac{e^{4x}}{e^x-2}.dx

Let

u^2 = e^x-2 \implies 2udu=e^x dx \implies dx = \dfrac{2u}{e^x}.du

So we have

\displaystyle \int_1^{\sqrt{2}} \dfrac{e^{4x}}{u^2} \cdot \frac{2u}{e^x} .du = 2\int_1^{\sqrt{2}} \frac{(u^2+2)^3}{u}.du = 2 \int_1^{\sqrt{2}} \frac{u^6+6u^4 +12 u^2 +8}{u}.du

I see exactly how that works, except for one thing- I don't know why the integral is equal to 2. Could you explain?

Oh, and how did you go from the third-to-last step to the penultimate step? I can see that you've binomially expanded to get the last step from the penultimate step, but I can't see how you got the fraction you have in the penultimate step in the first place.

(edited 6 years ago)

Reply 19