Integrating Logarithmic Functions (1 Viewer)

[twinklegal19]

I've tried rearranging it and using various trig identities, but I kept going around in circles and not getting it into any recognisable form. Wolfram Alpha didn't help either

 

[nightweaver066]

I've tried rearranging it and using various trig identities, but I kept going around in circles and not getting it into any recognisable form. Wolfram Alpha didn't help either

Following the question, divide numerator and denominator by cos^2x, so



Now using a suitable substitution.. then partial fractions..

 

[twinklegal19]

Following the question, divide numerator and denominator by cos^2x, so



Now using a suitable substitution.. then partial fractions..

But I only know 2U methods lol. Don't think substituting/partial fractions is needed

 

[twinklegal19]

Also not sure if it's a hard 2U question or easy 3U...

 

[RealiseNothing]

I don't think it's a 2U question, I can't think of how you would get in the answer given those limits.

 

[RealiseNothing]

Unless it ends up being something like:



Then by subbing in the limits you would get that answer.

 

[twinklegal19]

I don't think it's a 2U question, I can't think of how you would get in the answer given those limits.

Unless it ends up being something like:



Then by subbing in the limits you would get that answer.

ah ok, thanks

The source of this question isn't exactly known for strictly keeping questions 2U anyway. I'll just come back to it once we cover 3U methods then.

 

[Carrotsticks]

Bonus: Integrate that function, WITHOUT having first divided all terms by cos^2 (x).

 

[Sanjeet]

Bonus: Integrate that function, WITHOUT having first divided all terms by cos^2 (x).

But the questions says to

 

[twinklegal19]

But the questions says to

lol we will just have to trust him on this one

 

[Carrotsticks]

But the questions says to

I know it does, but I often promote 'alternative methods' to help build creativity with approaching problems (we are not sheep after all, are we?), which is often something valuable to have.

Here is an alternative way of doing the problem:

 

[twinklegal19]

Thanks! But how did you get from the first to second line? Partial fractions?

 

[Carrotsticks]

So you have to use partial fractions to get from the first to second line? Never used it for trig before

Well, you could consider it to be 'partial fractions' I guess, but you can easily 'guess' what is supposed to be in the numerator.

Here's what went on in my head:

1. The answer is ln(XXX), meaning that at some point I need to integrate something in the form f'(x)/f(x).

2. The only way I can get something in the form f'(x)/f(x) is if the top looks the same as the bottom, except with the functions swapped around (since derivative of sine is cos, and cos is - sine etc), so I just did that.

3. Oh look, a whole bunch of it cancels out if I cross multiply! But I also have a '6' lying around, so I'm going to chuck a 1/6 in front to 'balance' it out.

4. Good lord, it works out.

5. Add salt and pepper to season.

 

[twinklegal19]

Well, you could consider it to be 'partial fractions' I guess, but you can easily 'guess' what is supposed to be in the numerator.

Here's what went on in my head:

1. The answer is ln(XXX), meaning that at some point I need to integrate something in the form f'(x)/f(x).

2. The only way I can get something in the form f'(x)/f(x) is if the top looks the same as the bottom, except with the functions swapped around (since derivative of sine is cos, and cos is - sine etc), so I just did that.

3. Oh look, a whole bunch of it cancels out if I cross multiply! But I also have a '6' lying around, so I'm going to chuck a 1/6 in front to 'balance' it out.

4. Good lord, it works out.

5. Add salt and pepper to season.

1. Yeah, I kept trying to look for that form as well

2. oooohh I see now. I forgot how convenient it is with sine/cosine lol

3-4. Yay!

5. rofl

Thanks

 

[Sanjeet]

Yes but...I want to know how you can do it the way the question specifies. Anyone got a solution?

 

[D94]

Yes but...I want to know how you can do it the way the question specifies. Anyone got a solution?

Well, you could be smart and divide by cos²x then multiply by cos²x, then continue using Carrot's method

 

[Meshy590]

Yes but...I want to know how you can do it the way the question specifies. Anyone got a solution?

too hard mate

 

[D94]

Yes but...I want to know how you can do it the way the question specifies. Anyone got a solution?

Wait...do you want a 2U method or a 3/4U method?

If it's 3/4U, then after dividing, an appropriate substitution would be u = tan(x) and u' = sec², therefore, the integrand becomes 1/ (9 - u²) which partial fractions can be applied or another appropriate method.

 

[twinklegal19]

Well, you could be smart and divide by cos²x then multiply by cos²x, then continue using Carrot's method

ahahaha, true

Wait...do you want a 2U method or a 3/4U method?

If it's 3/4U, then after dividing, an appropriate substitution would be u = tan(x) and u' = sec², therefore, the integrand becomes 1/ (9 - u²) which partial fractions can be applied or another appropriate method.

yeah, that's was what nightweaver suggested before, and hopefully what I will learn soon.

 

[D94]

yeah, that's was what nightweaver suggested before, and hopefully what I will learn soon.

You should, I mean, it's essential. Anyway, knowing it now would be a decent head-start, especially so you can focus on other topics which you might find harder.