C4 Integration using standard patterns

krisshP

http://jonathanb.co.uk/cds/?q=C_4_6_E_2

I need help on part b please.

THANKS :smile:

Reply 1

CTArsenal

Use integration by substitution.

cosec^2(2x)cot(2x) is the same as cos(2x)/sin^3(2x).

Let u = sin(2x), then du/dx = 2cos(2x) -> du = 2cos(2x) dx -> 1/2 du = cos(2x) dx.

Then you have to integrate 1/2[u^(-3)] du, which is -u^(-2) / 4, which is -1/4u^2 and in terms of x that's -1/4sin^2(2x).

Which is equivalent to -cosec^2(2x)/4 + C.

(edited 10 years ago)

Reply 2

krisshP

Original post by CTArsenal

Use integration by substitution.

cosec^2(2x)cot(2x) is the same as cos(2x)/sin^3(2x).

Let u = sin(2x), then du/dx = cos(2x).

Then you have to integrate [u^(-3)] du, which is -u^(-2) / 2, which is -1/2u^2 and in terms of x that's -1/2sin^2(2x).

Which is equivalent to -cosec^2(2x)/2.

Yeah, but can I know why I went wrong above?

Reply 3

CTArsenal

Original post by krisshP

Yeah, but can I know why I went wrong above?

You've got the answer right :smile:

(1+cot^2(2x))/4 is the same as (cosec^2x)/4 using the trig identity of 1+cot^2(2x) = cosec^2(2x); but since the constant can take any value; it's absorbed into the answer. That's why we have different solutions to the actual solution.

(edited 10 years ago)

Reply 4

krisshP

Original post by CTArsenal

You've got the answer right :smile:

But then why is the solution different? They got cot^2(2x) instead of cosec^2(2x) :confused:

Edit: it's part b.

Reply 5

CTArsenal

Original post by krisshP

But then why is the solution different? They got cot^2(2x) instead of cosec^2(2x) :confused:

Edit: it's part b.

See edit above :smile:

The book generally does have some mistakes so don't always believe that they're perfect haha; but both answers are correct in this case.

Reply 6

krisshP

Original post by CTArsenal

See edit above :smile:

The book generally does have some mistakes so don't always believe that they're perfect haha; but both answers are correct in this case.

OH YEAH!!!! 1+cot^2(2x) =cosec^2(2x) MAKE SENSE NOW!!!!!!!! :biggrin:

THANKS A LOT! :smile:

Reply 7

krisshP

Original post by CTArsenal

WOW! It worked as expected! How nice?

Did you decide what u was equal to after realising that cosec^2(2x)cot(2x) is equal to cos(2x)/sin^3(2x)?

Thanks

Reply 8

CTArsenal

Original post by krisshP

WOW! It worked as expected! How nice?

Did you decide what u was equal to after realising that cosec^2(2x)cot(2x) is equal to cos(2x)/sin^3(2x)?

Thanks

I somewhat have a knack for seeing substitutions for integration (if they're possible) haha. I would just say that if you're stuck with a question like that in the exam then just write it out a different way and see if that makes the question any easier :smile:

Reply 9

krisshP

Original post by CTArsenal

I'm guessing you let u=sin(2x) as then inside the integral you'd then have u raised to integer powers which are easier to deal with. Also, u=sin(2x) differentiates to 2cos(2x) which ends up cancelling quite nicely.

Thanks for that tip.

Reply 10

krisshP

Original post by CTArsenal

For the last part there (part e) what would you let u be equal to and why?

Thanks

Edit: I let u=cosx and I ended up with -2S (x^2)/(u^3) du.

(edited 10 years ago)

Reply 11

CTArsenal

Original post by krisshP

Edit: I let u=cosx and I ended up with -2S (x^2)/(u^3) du.

You wouldn't be able to integrate that by substitution because of the x^2, I'd do integration by parts, and let 2x^2 = u, and dv/dx = sec^2(x)tan(x).

Integrate sec^2(x)tan(x) using substitution separately, as it's the same as sin(x)/cos^3(x). Then let a variable equal cos x and then integrate. When you get that solution then it's just 2x^2 x (your solution) - the integral of 4x x (your solution), and repeat the process.

Reply 12

davros

Original post by krisshP

Just tried it and didn't go well at all. I did it, but at the bold part things just get so annoying and confused.

You do need 2 iterations of IBP to make the last example work.

This is a skill that you will need to get used to, so I would suggest starting with something a bit simpler like the integral of (x^2)e^x or (x^2)sin x and making sure you can integrate those by parts first (check your answer by differentiating it!!).

Once you're happy with the process then you can try out a more complicated problem like your last example :smile:

Reply 13

krisshP

Original post by davros

OK, thanks

Reply 14

krisshP

Original post by davros

Yay done it. It actually did help to go back to simpler ones and then come back to this beast as it made me select u in by parts more appropriately which made it significantly easier! :smile:

THANKS