C4 Integration (Edexcel)

Would much appreciate some help on this question:

Integrate:

{sin}^2{x} {cos}^2{x}

Using trig functions/identities involving cos 2x, I got:

\int (\frac{1}{2} - \frac{1}{2}cos 2x)(\frac{1}{2}cos 2x + \frac{1}{2})\ dx[br][br][br][br][br]= \int \frac{1}{4} - \frac{1}{4}{cos}^2{2x}\ dx[br][br][br][br][br]= \int \frac{1}{4} - \frac{1}{4}(\frac{1}{2}cos 2x + \frac{1}{2})\ dx[br][br][br][br][br]= \int \frac{1}{4} - \frac{1}{8} cos 2x - \frac{1}{8}\ dx = \frac{1}{8}x - \frac{1}{16}sin 2x + c

......but they got

- \frac{1}{32} sin2x

instead of

- \frac{1}{16} sin 2x

for the final answer?

I think I might have gone wrong on the third line with the substitution of

{cos}^2{2x}

?

Also, how do you integrate

\frac{1}{{sin}^2{x} {cos}^2{x}}

Reply 1

soutioirsim

sinAcosA = \frac{1}{2}sin2x

This should help.

EDIT: What the above post said basically ^^

Reply 2

Introverted moron

Original post by soutioirsim

sinAcosA = \frac{1}{2}sin2x

This should help.

EDIT: What the above post said basically ^^

For integrating the reciprocal, you mean?

Reply 3

Phil1541

I suck with Latex but I'll give it a go

I got x/8 - Sin4x / 32 +C :eek:

I used (1/2 sin2x)^2

Expanded then found 1/4 Sin^2 2x in terms of cos4x.

I think I may have made a mistake aswell somewhere :frown:

is something wrong with the 1/4 Sin^2 2x into terms of cos4x ?

Reply 4

soutioirsim

Original post by Introverted moron

For integrating the reciprocal, you mean?

Yeah, so you have:

\frac{1}{(\frac{1}{2}sin2x)^2}

Can you see what to do from there?

Reply 5

Introverted moron

Original post by soutioirsim

Yeah, so you have:

\frac{1}{(\frac{1}{2}sin2x)^2}

Can you see what to do from there?

Convert that to

Unparseable latex formula:

}{(\frac{1}{2}sin2x)^{-2}

and use the chain rule? :erm:

Reply 6

gbsn350

i found when you got to integral 1/4 - 1/4(cos^2 2x) dx...

when you replace the (cos^2 2x) with (1/2cos2x + 1/2) it should be (1/2cos4x +1/2) so then when you divide by differential should give 1/32?

Reply 7

gbsn350

sorry didnt really explain. you replaced it with the double angle identity as if it was just (cos^2 x) when in this case it was (cos^2 2x) so you had to multiply the x by 2.

Reply 8

Farhan.Hanif93

Original post by Introverted moron

Convert that to

Unparseable latex formula:

}{(\frac{1}{2}sin2x)^{-2}

and use the chain rule? :erm:

You can't use the chain rule because this isn't differentiation. If you were referring to the reverse chain rule, you can't use that either because the integrand isn't of the required form. Instead notice that

\dfrac{1}{\sin ^2 (2x)} \equiv cosec ^2 (2x)

, which is a standard integral.

Reply 9

gbsn350

just gone through it again and not totally sure thats right, because that would give the final answer 1/8x - 1/32 sin4x +c which doesn't sound right....

Reply 10

Introverted moron

Original post by gbsn350

sorry didnt really explain. you replaced it with the double angle identity as if it was just (cos^2 x) when in this case it was (cos^2 2x) so you had to multiply the x by 2.

So if

cos 2x = 2{cos}^{2}x - 1

, then

2{cos}^{2}2x - 1

must equal

cos 4x

Reply 11

Phil1541

Original post by Introverted moron

Convert that to

Unparseable latex formula:

}{(\frac{1}{2}sin2x)^{-2}

and use the chain rule? :erm:

Your book has the answer wrong I just checked with wolfram alpha and the actual answer is:

x/8 - 1/32 sin4x +C

(for the first question)

For the second, I would write 1/ (1/4 Sin^2 2x) as 4cosec^2 2x

Then you can integrate 4 cosec^2 2x to become -2cot2x using the formula in your formula booklets.

Integral Cosec^2 x = -cotx

Hope this helps :smile:

Wolfram alpha is really useful to check your answers with

http://www.wolframalpha.com/input/?i=1%2F%28sin%5E2x+cos%5E2x%29+integrate

Reply 12

Phil1541

Original post by gbsn350

just gone through it again and not totally sure thats right, because that would give the final answer 1/8x - 1/32 sin4x +c which doesn't sound right....

Thats the correct answer the book must be wrong, just checked it up on wolfram alpha :smile:

Reply 13

Introverted moron

Original post by Farhan.Hanif93

\dfrac{1}{\sin ^2 (2x)} \equiv cosec ^2 (2x)

, which is a standard integral.

Ah, I see. I'll have a go at it later on/tomorrow using your pointers; I know it's not difficult from here, but my brain's a little fried at the moment. :redface:

Thank you and everyone else who has posted for your help.

Reply 14

Introverted moron

Original post by Phil1541

Thats the correct answer the book must be wrong, just checked it up on wolfram alpha :smile:

No, the book's right. I'm wrong. It is sin 4x. :biggrin:

Reply 15

gbsn350

basically what im saying is... at the beginning you subbed in (1/2 cos2x + 1/2) for cos^2 x.

then on the third line you've got cos^2 2x but you've subbed in (1/2 cos2x +1/2) again which can't be right because its a 2x not just an x.

so you should of subbed in (1/2 cos4x + 1/2)

Reply 16

Introverted moron

Original post by gbsn350

Yeah, I can see the logic behind that now. Thanks for your help.

Reply 17

tiny hobbit

Original post by Introverted moron

Yeah, I can see the logic behind that now. Thanks for your help.

It is often useful when to integrating (1/something) to bring it onto the top. With trig it will often have a different name. With e^x you will get a negative power.