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I need help with some C4 integration

How do I integrate this:
sin2xcos2xsin^2 xcos^2 x

what I've done so far is as follows:
(sinxcosx)2(sinxcosx)^2

(12sin2x)2(\frac{1}{2}sin2x)^2

12sin22x\frac{1}{2}sin^2 2x

since sin2xsin^2 x = 12(1cos2x)\frac{1}{2}(1 - cos2x)

12sin22x\frac{1}{2}sin^2 2x = 14(1cos4x)\frac{1}{4}(1 - cos4x)

1414cos4x\frac{1}{4} - \frac{1}{4}cos4x

integrate 14\frac{1}{4} = 14x\frac{1}{4} x

integrate 14cos4x\frac{1}{4} cos4x : let y = 14sin4x\frac{1}{4} sin4x

so dy/dx = cos4x, therefore the integral of 14cos4x\frac{1}{4} cos4x = 116sin4x\frac{1}{16} sin4x

my final answer = 14x116sin4x+c\frac{1}{4} x - \frac{1}{16} sin4x + c

but the answer book says its supposed to be: 18x132sin4x+c\frac{1}{8} x - \frac{1}{32} sin4x + c

can someone tell me where i've gone wrong please....
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I've also got this to integrate: (secxtanx)^2
But I have no idea how to do it.........
Reply 1
(12sin2x)2(\frac{1}{2}sin2x)^{2} isn't the same as (12sin22x) (\frac{1}{2}sin^{2}2x) you need to square the 12\frac{1}{2} as well.
W.H.T
How do I integrate this:
sin2xcos2xsin^2 xcos^2 x

what I've done so far is as follows:
(sinxcosx)2(sinxcosx)^2

(12sin2x)2(\frac{1}{2}sin2x)^2

12sin22x\frac{1}{2}sin^2 2x

since sin2xsin^2 x = 12(1cos2x)\frac{1}{2}(1 - cos2x)

12sin22x\frac{1}{2}sin^2 2x = 14(1cos4x)\frac{1}{4}(1 - cos4x)

1414cos4x\frac{1}{4} - \frac{1}{4}cos4x

integrate 14\frac{1}{4} = 14x\frac{1}{4} x

integrate 14cos4x\frac{1}{4} cos4x : let y = 14sin4x\frac{1}{4} sin4x

so dy/dx = cos4x, therefore the integral of 14cos4x\frac{1}{4} cos4x = 116sin4x\frac{1}{16} sin4x

my final answer = 14x116sin4x+c\frac{1}{4} x - \frac{1}{16} sin4x + c

but the answer book says its supposed to be: 18x132sin4x+c\frac{1}{8} x - \frac{1}{32} sin4x + c

can someone tell me where i've gone wrong please....
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I've also got this to integrate: (secxtanx)^2
But I have no idea how to do it.........


Nothing major - you just missed a factor of 1/2:

(12sin2x)2=14sin2(2x)(\frac{1}{2}sin2x)^2 = \mathbf{\dfrac{1}{4}}sin^2(2x)

As for (sec x tan x)^2, try using the substitution

u = tan x
Reply 3
so for (sec x tan x)^2, would integration by parts not work?

I thought that as 'secxtanx' is a product, that I should use the reverse of the product rule for differentiation which would be integration by parts
Reply 4
Cos^2xsin^2x

Cos2x= 1 -2sin^2x
(Cos2x-1)/-2 =sin^2x
-1/2cos2x + 1/2 =sin^2x
Also
Cos2x= 2cos^2x-1
(Cos2x+1)/2=cos^2x
1/2cos2x + 1/2 substitute
(1/2cos2x + 1/2)x( -1/2cos2x + 1/2)
1/2cos2x(-1/2cos2x + ½) + 1/2(-1/2cos2x + 1/2)
-1/4cos^2(2x) + 1/4cos2x -1/2cos2x + ¼
Cos4x= 2 cos^2(2x)-1
(Cos4x+1)/2
= 1/2cos4x +1/2
So -1/4(1/2cos4x +1/2) +1/4
-1/8cos4x -1/8 + ¼
Integrate: -1/8cos4x + 1/8
integrate -1/8cos4x first
Y= -1/8sin4x
U= 4x so y=-1/8sinu
du/dx= 4 and
dy/du= -1/8cosu
so dy/dx=-1/2cos4x
so its: -1/32sin4x + c
integrate 1/8 to get 1/8x
so ans: 1/8x -1/32sin4x + c

I hope it makes sense
Reply 5
F.I
I hope it makes sense


Learn Latex. Now.
Reply 6
F.I
Cos^2xsin^2x

Cos2x= 1 -2sin^2x
(Cos2x-1)/-2 =sin^2x
-1/2cos2x + 1/2 =sin^2x
Also
Cos2x= 2cos^2x-1
(Cos2x+1)/2=cos^2x
1/2cos2x + 1/2 substitute
(1/2cos2x + 1/2)x( -1/2cos2x + 1/2)
1/2cos2x(-1/2cos2x + ½) + 1/2(-1/2cos2x + 1/2)
-1/4cos^2(2x) + 1/4cos2x -1/2cos2x + ¼
Cos4x= 2 cos^2(2x)-1
(Cos4x+1)/2
= 1/2cos4x +1/2
So -1/4(1/2cos4x +1/2) +1/4
-1/8cos4x -1/8 + ¼
Integrate: -1/8cos4x + 1/8
integrate -1/8cos4x first
Y= -1/8sin4x
U= 4x so y=-1/8sinu
du/dx= 4 and
dy/du= -1/8cosu
so dy/dx=-1/2cos4x
so its: -1/32sin4x + c
integrate 1/8 to get 1/8x
so ans: 1/8x -1/32sin4x + c

I hope it makes sense


Thanks

by the way, do you know how I could integrate (secxtanx)^2
Reply 7
W.H.T
Thanks

by the way, do you know how I could integrate (secxtanx)^2

u=secx ought to do the trick
Reply 8
Pheylan
u=secx ought to do the trick


How would I sub in 'u = secx' ?

(secxtanx)^2

would I change it to make: sec^2 x * tan^2 x ?

I'm not too familar with substitution method, can you help me out a bit more please?
Reply 9
opps sorry use tan^3x

so y=tan^3x
y=(tanx)^3
dy/dx=3(tanx)^2 x sec^2x

1/3tan^3x + c
Reply 10
F.I
opps sorry use tan^3x

so y=tan^3x
y=(tanx)^3
dy/dx=3(tanx)^2 x sec^2x

1/3tan^3x + c



wow , so I don't need to use substitution ?

but just out of interest, if I did you substitution as my method to integrate (secxtanx)^2 as others have suggested, would I still have got to the answer of '1/3tan^3x + c' ?
Reply 11
W.H.T
wow , so I don't need to use substitution ?

but just out of interest, if I did you substitution as my method to integrate (secxtanx)^2 as others have suggested, would I still have got to the answer of '1/3tan^3x + c' ?


i dont think you substitute otherwise you get to the power of 4 which is difficult to integrate...this is the method i know

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