c4 integration help!

quirksy

questions 3 and 4 - any ideas?

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What have you tried?

Original post by quirksy

questions 3 and 4 - any ideas?

different techniques required for most of them ...

Reply 3

quirksy

Original post by zetamcfc

What have you tried?

Original post by TeeEm

different techniques required for most of them ...

well i'm currently attempting 3a, and trying integration by substitution...

substituted in x=3sin(u) to get integral of 1/(3-3sin(u)) 3 cos(u) du.

Reply 4

TeeEm

Original post by quirksy

well i'm currently attempting 3a, and trying integration by substitution...

substituted in x=3sin(u) to get integral of 1/(3-3sin(u)) 3 cos(u) du.

you mean1/(9-9sin²(u)) 3 cos(u) du

and also a square root

factorize the 9 under the square root ...

Reply 5

Protoxylic

Original post by quirksy

well i'm currently attempting 3a, and trying integration by substitution...

substituted in x=3sin(u) to get integral of 1/(3-3sin(u)) 3 cos(u) du.

If you substituted x=3sin(u) then x^2 wouldn't be 3sin(u). You should have integral 1/sqrt[(9-9sin^2(u)] . 3cos(u) du

You have said that sqrt(9-9x^2)=3-3x which isn't true.

Reply 6

TeeEm

too many of us here ...
I am the weakest link...
goodbye

Reply 7

quirksy

Original post by TeeEm

you mean1/(9-9sin²(u)) 3 cos(u) du

and also a square root

factorize the 9 under the square root ...

But I square rooted the 9 - 9sin²(u) to get 3 - 3sin(u)?

Reply 8

TeeEm

Original post by quirksy

But I square rooted the 9 - 9sin²(u) to get 3 - 3sin(u)?

please read post 6 and post 7

Reply 9

davros

Original post by quirksy

But I square rooted the 9 - 9sin²(u) to get 3 - 3sin(u)?

That's not how square roots work!

Are you saying that

\sqrt{a^2 - b^2} = a - b

, because that isn't true at all!!

Reply 10

quirksy

Original post by Protoxylic

If you substituted x=3sin(u) then x^2 wouldn't be 3sin(u). You should have integral 1/sqrt[(9-9sin^2(u)] . 3cos(u) du

You have said that sqrt(9-9x^2)=3-3x which isn't true.

oh right, thank you! is this correct?

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Reply 11

Protoxylic

Original post by quirksy

oh right, thank you! is this correct?

Correct

Reply 12

Hasufel

i find these ones interesting, so i`ll indicate them:

4).

b) use the substitution:

u=x^{2}+2x

d) use the relation:

Unparseable latex formula:

\displaystyle \sinA \cosB = \frac{1}{2} ( \sin(A+B) + \sin(A-B))

to reduce to a more integrable form.

f) divide out into partial fraction then use logs to integrate.

I) use integration by parts (or tabular integration - it`s quicker)

j) expand out, and reduce by using the identity:

- \cot^{2}(x)=1- \csc^{2}(x)

k) use partial fractions.

(edited 9 years ago)

Reply 13

MutlohJ

It really is quite simple, have you tried using examsolutions.net the guy doing the videos is extremely helpful.

Reply 14

TenOfThem

Original post by quirksy

oh right, thank you! is this correct?

No

You need to change your limits

Reply 15

TenOfThem

Original post by buxtonarmy

How'd you solve 4c?

You turn it into A + B/(2x+1)

Reply 16

davros

Original post by buxtonarmy

What does that mean?

It means what it says!

You can rewrite the integrand as the sum of a constant and a fraction which has a number on top and a linear factor on the bottom.

You know how to integrate a constant and you should know how to integrate something of the form A/(ax + b) where A, a and b are constants :smile: