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###### Help! Hyperbolic integration int (cosh^5(2x)sinh(2x)) dx

1 year ago

HOW DO I DO THIS QUESTION??

I tried making it so (cosh(2x))^5 × (sinh(2x))

And then integrating each separately so 1/6 × 1/2 sinh(2x) + 1/2cosh(2x)

But it doesn't work obviously but I'm not sure how to do it and the test book skips so many stages.

I tried making it so (cosh(2x))^5 × (sinh(2x))

And then integrating each separately so 1/6 × 1/2 sinh(2x) + 1/2cosh(2x)

But it doesn't work obviously but I'm not sure how to do it and the test book skips so many stages.

Original post by danai m

You have to do this:

1) Let y = cosh^6(2x)

2) Find dy/dx ( = 6cosh^5(2x) x 2sinh(2x) = 12sinh(2x)cosh^5(2x))

3) Compare dy/dx with the integral

4) Divide cosh^6(2x) by 12

3) Compare

1) Let y = cosh^6(2x)

2) Find dy/dx ( = 6cosh^5(2x) x 2sinh(2x) = 12sinh(2x)cosh^5(2x))

3) Compare dy/dx with the integral

4) Divide cosh^6(2x) by 12

3) Compare

Could you write it on paper mayhaps? Also what's this method of integrating called ??

Original post by FloryK

Could you write it on paper mayhaps? Also what's this method of integrating called ??

This is really no different from integrating something like $\sin^5{x} \cos{x}$ - when you're familiar with it you can spot it by recognition but otherwise a substitution like y = sin x will do the trick. You must have come across this technique already?

Original post by davros

This is really no different from integrating something like $\sin^5{x} \cos{x}$ - when you're familiar with it you can spot it by recognition but otherwise a substitution like y = sin x will do the trick. You must have come across this technique already?

I hadn't realised there was a response sorry! I'm really fuzzy on this topic so not certain if I have come across it or not, could you write that out on paper?

Original post by FloryK

I hadn't realised there was a response sorry! I'm really fuzzy on this topic so not certain if I have come across it or not, could you write that out on paper?

We can't give solutions as that is against the rules of the forum, but perhaps it would help if you went over what integration techniques you've covered. Presumably you've done integration by substitution before?

Original post by davros

We can't give solutions as that is against the rules of the forum, but perhaps it would help if you went over what integration techniques you've covered. Presumably you've done integration by substitution before?

No I'm not asking for the solution I'm simply asking for a WORKED example, why is this always the response an example //= solution !! 🥲

Yes I have done PBS but just worked solutions generally work very well in helping people so I thought I'd ask but it's fine as this thread is quite old I'm not even sure where I put my paper anymore ughhhhhubw

Original post by FloryK

No I'm not asking for the solution I'm simply asking for a WORKED example, why is this always the response an example //= solution !! 🥲

Yes I have done PBS but just worked solutions generally work very well in helping people so I thought I'd ask but it's fine as this thread is quite old I'm not even sure where I put my paper anymore ughhhhhubw

Yes I have done PBS but just worked solutions generally work very well in helping people so I thought I'd ask but it's fine as this thread is quite old I'm not even sure where I put my paper anymore ughhhhhubw

You wrote "could you write that out on paper" so I assumed you specifically wanted a worked solution

As a similar example, suppose you were integrating ((ln x)^5 )/ x with respect to x, If you set u = ln x then du = dx / x so you're integrating u^5 with respect to u - final answer (u^6) / 6 + k or in terms of the original function ((ln x)^6) / 6 + k.

You must have loads of similar examples in your textbook if you look under "Integration by substitution"

Original post by davros

You wrote "could you write that out on paper" so I assumed you specifically wanted a worked solution

As a similar example, suppose you were integrating ((ln x)^5 )/ x with respect to x, If you set u = ln x then du = dx / x so you're integrating u^5 with respect to u - final answer (u^6) / 6 + k or in terms of the original function ((ln x)^6) / 6 + k.

You must have loads of similar examples in your textbook if you look under "Integration by substitution"

As a similar example, suppose you were integrating ((ln x)^5 )/ x with respect to x, If you set u = ln x then du = dx / x so you're integrating u^5 with respect to u - final answer (u^6) / 6 + k or in terms of the original function ((ln x)^6) / 6 + k.

You must have loads of similar examples in your textbook if you look under "Integration by substitution"

Yes don't worry, the textbook isn't very helpful as the initial question is from the text book but it's okay! I'll just figure it out at some point later, I may just need to look over it and realise it's not too bad but for now ignoring it haha.

Anyways thank you!

Original post by FloryK

Could you write it on paper mayhaps? Also what's this method of integrating called ??

The A-level textbook calls it "reverse chain rule", the more formal term is "integration by inspection" - i.e. looking at it and having a guess. However this is all just integration by substitution but skipping a few steps.

Original post by Sinnoh

The A-level textbook calls it "reverse chain rule", the more formal term is "integration by inspection" - i.e. looking at it and having a guess. However this is all just integration by substitution but skipping a few steps.

Do they actually use this terminology in the textbooks now? Because (a) it's a horrible expression that never used to be used in calculus texts, and (b) it seems to cause immense confusion for students who quote it on TSR because they often think they can apply it in all sorts of situations when it just isn't applicable and start dividing by derivatives all over the place and getting horribly wrong answers!

"Recognition" / "inspection" comes with experience, but for the learner substitution is almost always the "safer" way to go and is a standard integration technique

Original post by davros

Do they actually use this terminology in the textbooks now? Because (a) it's a horrible expression that never used to be used in calculus texts, and (b) it seems to cause immense confusion for students who quote it on TSR because they often think they can apply it in all sorts of situations when it just isn't applicable and start dividing by derivatives all over the place and getting horribly wrong answers!

"Recognition" / "inspection" comes with experience, but for the learner substitution is almost always the "safer" way to go and is a standard integration technique

"Recognition" / "inspection" comes with experience, but for the learner substitution is almost always the "safer" way to go and is a standard integration technique

Yeah that's the name of the sub-chapter. Don't know what they were thinking, it contributes to making maths seem like all you do is learn a bunch of "tricks" to answer questions. Back when I was doing A-levels and helping people with the topic I always tried to emphasise to them that this is all really just integration by substitution.

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