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C4 Integration help please

Could someone help me with this question please:

How would i integrate sin2xcos2x sin^2xcos^2x?
This is Question 1h Exercise 6C from C4 book.
.....................................................................
(edited 11 years ago)

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Reply 1
Avatar for nerak99
nerak99
13
Well you do need to use double angle formula but you can just reapply to deal with cos⁡22x\cos^2 2x

cos⁡4x=2cos⁡22x−1\cos 4x = 2 \cos^2 2x -1
(edited 11 years ago)
Original post by nm786
Could someone help me with this question please:

How would i integrate sin2xcos2x sin^2xcos^2x?
This is Question 1h Exercise 6C from C4 book.

This is what i did:
cos2x=1−2sin2xcos2x=1-2sin^2x and cos2x=2cos2x−1cos2x=2cos^2x-1
So, sin2x=−1/2(cos2x−1)sin^2x=-1/2(cos2x-1) and cos2x=1/2(cos2x+1)cos^2x=1/2(cos2x+1)
multiplying both of these: −1/4(cos22x−1)-1/4(cos^22x-1)
what should i do next? I'm confused.


sin⁡2xcos⁡2x=sin⁡22x4\displaystyle \sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}
Reply 3
Avatar for nm786
nm786
OP
2
Original post by nerak99
Well you do need to use double angle formula but you can just reapply to deal with cos⁡22x\cos^2 2x

cos⁡4x=2cos⁡22x−1\cos 4x = 2 \cos^2 2x -1

Thanks for replying mate, so sould i sub this into the last equation and and integrate?: 1/2(1/2(cos4x+1)+1)
Reply 4
Avatar for nerak99
nerak99
13
Well, working from what you did you could do that (I have not checked your working) but what Mr M showed you is a more elegant way of working if you want to take it from the top.
You need to use cos⁡4x=1−2sin⁡22x\cos 4x = 1 - 2 \sin^2 2x

Mr M has use sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x and squared it.
(edited 11 years ago)
Reply 5
Avatar for nm786
nm786
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2
Original post by nerak99
Well, working from what you did you could do that (I have not checked your working) but what Mr M showed you is a more elegant way of working if you want to take it from the top.
You need to use cos⁡4x=1−2sin⁡22x\cos 4x = 1 - 2 \sin^2 2x

Mr M has use sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x and squared it.

so from what i did: 1/2(1/2(cos4x+1)+1)
=1/2(1/2cos4x+1/2+1)
=1/4cos4x+1/4+1/2
integrating above = 1/4*1/4sin4x+1/4x+1/2x+c
=1/16sin4x+3/4x+c
but the answer is 1/8x-1/32sin4x+c?

I'll use the equation Mr M suggested in a minute.
(edited 11 years ago)
Reply 6
Avatar for nm786
nm786
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Original post by Mr M
sin⁡2xcos⁡2x=sin⁡22x4\displaystyle \sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}

I understand how you got that now but how would i go about integrating it?
Original post by nm786
I understand how you got that now but how would i go about integrating it?


You should know that sin⁡2θ=1−cos⁡2θ2\displaystyle \sin^2 \theta = \frac{1-\cos 2\theta}{2}
Reply 8
Avatar for nm786
nm786
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Original post by Mr M
You should know that sin⁡2θ=1−cos⁡2θ2\displaystyle \sin^2 \theta = \frac{1-\cos 2\theta}{2}

i never knew that :frown:
what about cos^2x?
Original post by nm786
i never knew that :frown:
what about cos^2x?


It is a key fact in Core 4 integration.

cos⁡2θ=1+cos⁡2θ2\displaystyle \cos^2 \theta= \frac{1 + \cos 2\theta}{2} but that is not relevant to this question.
Reply 10
Avatar for nm786
nm786
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Original post by Mr M
It is a key fact in Core 4 integration.

cos⁡2θ=1+cos⁡2θ2\displaystyle \cos^2 \theta= \frac{1 + \cos 2\theta}{2} but that is not relevant to this question.

alright i get it this is from cos2x identity right?
(edited 11 years ago)
Original post by nm786
alright i get it this is from cos2x identity right?
how would i do this question sir,
using what you said: sin22x/4=sin2xcos2xsin^22x/4=sin^2xcos^2x
=(2sinxcosx)2=1/2sin2xcos2x=(2sinxcosx)^2=1/2sin^2xcos^2x
using the identitiessin2x=−1/2(cos2x−1)sin^2x=-1/2(cos2x-1) and cos2x=1/2(cos2x+1)cos^2x=1/2(cos2x+1) multiplying them together and 1/2:
=(1/2)[(−1/4(cos22x−1)]=(1/2)[(-1/4(cos^22x-1)]
=−1/8cos22x+1/8=-1/8cos^22x+1/8
I can use −1/2(cos2x−1)againso−1/8(−1/2cos2x−1)+1/8-1/2(cos2x-1) again so -1/8(-1/2cos2x-1)+1/8
=1/16cos2x+1/8+1/8=116cos2x+2/8=1/16cos2x+1/8+1/8=116cos2x+2/8
integrating above i get (1/2)(1/16)sin2x+2/8x(1/2)(1/16)sin2x+2/8x i have no idea where i went wrong :s-smilie:


You make things so complicated!

∫sin⁡2xcos⁡2x dx=∫sin⁡22x4 dx=∫1−cos⁡4x8 dx\displaystyle \int \sin^2 x \cos^2 x \, dx = \int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx
Reply 12
Avatar for nm786
nm786
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2
Original post by Mr M
You make things so complicated!

∫sin⁡2xcos⁡2x dx=∫sin⁡22x4 dx=∫1−cos⁡4x8 dx\displaystyle \int \sin^2 x \cos^2 x \, dx = \int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx

:colondollar: i don't understand how you went from ∫sin⁡22x4 dx=∫1−cos⁡4x8 dx\int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx
Original post by nm786
:colondollar: i don't understand how you went from ∫sin⁡22x4 dx=∫1−cos⁡4x8 dx\int \frac{\sin^2 2x}{4} \, dx = \int \frac{1-\cos 4x}{8} \, dx


Post 9?
Reply 14
Avatar for nm786
nm786
OP
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Original post by Mr M
Post 9?

oh yeah sorry,
so ∫1−cos⁡4x8 dx=∫1/8−cos4x8 dx \int \frac{1-\cos 4x}{8} \, dx = \int 1/8-\frac{cos4x}{8} \, dx
=1/8−(1/8)(1/4)sin4x+c=1/8-(1/8)(1/4)sin4x+c
EDIT:=1/8x−1/32sin4x+C=1/8x-1/32sin4x+C which is the right answer!
Cheers Sir! :smile:
(edited 11 years ago)
Original post by nm786
oh yeah sorry,
so ∫1−cos⁡4x8 dx=∫1/8−cos4x8 dx \int \frac{1-\cos 4x}{8} \, dx = \int 1/8-\frac{cos4x}{8} \, dx
=1/8−(1/8)(1/4)sin4x+c=1/8-(1/8)(1/4)sin4x+c
=1/8−1/32sin4x+C=1/8-1/32sin4x+C which is the right answer!
Cheers Sir! :smile:


∫18 dx=x8+k\displaystyle \int \frac{1}{8} \, dx = \frac{x}{8}+k
Reply 16
Avatar for nm786
nm786
OP
2
Original post by Mr M
∫18 dx=x8+k\displaystyle \int \frac{1}{8} \, dx = \frac{x}{8}+k

damn it i forgot the x :colondollar:
Much appreciated for the help sir!
Original post by nm786
damn it i forgot the x :colondollar:
Much appreciated for the help sir!


You are welcome.
Reply 18
Avatar for ztibor
ztibor
10
Original post by nm786
alright i get it this is from cos2x identity right?
how would i do this question then sir,
using what you said: sin22x/4=sin2xcos2xsin^22x/4=sin^2xcos^2x
=(2sinxcosx)2=1/2sin2xcos2x=(2sinxcosx)^2=1/2sin^2xcos^2x

do you think to
(2sin⁡xcos⁡x)24=4⋅sin⁡2xcos⁡2x4=sin⁡2x⋅cos⁡2x\frac{(2\sin x\cos x)^2}{4}=\frac{4\cdot \sin^2 x \cos^2 x}{4}=\sin^2 x\cdot \cos^2 x
here

using the identitiessin2x=−1/2(cos2x−1)sin^2x=-1/2(cos2x-1) and cos2x=1/2(cos2x+1)cos^2x=1/2(cos2x+1) multiplying them together and 1/2:

There would not be 'and 1/2'

=(1/2)[(−1/4(cos22x−1)]=(1/2)[(-1/4(cos^22x-1)]
=−1/8cos22x+1/8=-1/8cos^22x+1/8

That is without 1/2 factor
1−cos⁡22x4=sin⁡22x4\frac{1-\cos^2 2x}{4}=\frac{\sin^2 2x}{4}
which is same that was said to you.

I can use −1/2(cos2x−1)-1/2(cos2x-1) again so −1/8(−1/2cos2x−1)+1/8-1/8(-1/2cos2x-1)+1/8

Just little difference only
Consider that cos⁡(2⋅2x)=cos⁡4x \cos (2\cdot 2x)=\cos 4x
(and +1/4 at the end)
1−cos⁡22x4=14−14cos⁡22x=\frac{1-\cos^2 2x}{4}=\frac{1}{4}-\frac{1}{4}\cos^2 2x=
=14−18(1+cos⁡4x)=18−18cos⁡4x=\frac{1}{4}-\frac{1}{8}(1+\cos 4x)=\frac{1}{8}-\frac{1}{8}\cos 4x

=1/16cos2x+1/8+1/8=116cos2x+2/8=1/16cos2x+1/8+1/8=116cos2x+2/8
integrating above i get (1/2)(1/16)sin2x+2/8x+C(1/2)(1/16)sin2x+2/8x +C i have no idea where i went wrong the answer is something else :s-smilie:
Original post by ztibor
do you think to
(2sin⁡xcos⁡x)24=4⋅sin⁡2xcos⁡2x4=sin⁡2x⋅cos⁡2x\frac{(2\sin x\cos x)^2}{4}=\frac{4\cdot \sin^2 x \cos^2 x}{4}=\sin^2 x\cdot \cos^2 x
here

There would not be 'and 1/2'

That is without 1/2 factor
1−cos⁡22x4=sin⁡22x4\frac{1-\cos^2 2x}{4}=\frac{\sin^2 2x}{4}
which is same that was said to you.

Just little difference only
Consider that cos⁡(2⋅2x)=cos⁡4x \cos (2\cdot 2x)=\cos 4x
(and +1/4 at the end)
1−cos⁡22x4=14−14cos⁡22x=\frac{1-\cos^2 2x}{4}=\frac{1}{4}-\frac{1}{4}\cos^2 2x=
=14−18(1+cos⁡4x)=18−18cos⁡4x=\frac{1}{4}-\frac{1}{8}(1+\cos 4x)=\frac{1}{8}-\frac{1}{8}\cos 4x


He finished the question 25 minutes ago.

:s-smilie:

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