Integrating Sin^2xCos^2x (1 Viewer)

[iptteacherssuck]

How do you integrate Sin^2xCos^2x

 

[BlackJack]

use cos 2x= 2cos^2x - 1
cos 2x= 1 - 2sin^2x
Integral is:
1/4 (1 - cos^2 (2x) )
continue from there...

 

[wogboy]

since sin[2a] = 2 * sin[a] * cos [a]

therefore,

integral(sin^2[x] * cos^2[x]) dx

= integral(1/4 * sin^2[2x]) dx

= 1/4 * integral(sin^2[2x]) dx

now since,

sin^2[2x] = (1 - cos[4x]) / 2

therefore,

= 1/2 * 1/4 * integral(1 - cos[4x]) dx
= 1/2 * 1/4 * integral(1 - cos[4x]) dx

= x/8 - (sin[2x])/32 + C

[Edit] Minor correction as pointed out by BJ.

 

[BlackJack]

Originally posted by wogboy

sin^2[2x] = (1 - cos[4x]) / 2)

minor slip-up:
sin^2(2x) = (1 - cos[4x]) /2
then it goes to
= 1/4 * integral(0.5*(1 - cos[4x]) dx
= 1/8 * integral(1 - cos[4x]) dx
= x/8 - (sin[2x])/32 + C

We did the question elsewhere w/ the cosine rule instead...

 

[addikaye03]

int. sin^2xcos^2x

int. 1/4(2sinxcosx)^2

1/4 int. sin^2 (2x)

[cos4x=1-2sin^2(2x)]

1/8 int. sin^2(2x)=1/8 int. (1-cos4x)

1/8(x-1/4sin4x)+C #

 

[edmund gosse]

I get x/8 + 1/32(sin4x) + c

 

[addikaye03]

I get x/8 + 1/32(sin4x) + c

thats the answer dude, i got the same answer, just expand bracket.

the girl above me stuffed it up, but since this was created in 2002 and they would be 26-27 years old, i don't think its gona matter

 

[edmund gosse]

thats the answer dude, i got the same answer, just expand bracket.

the girl above me stuffed it up, but since this was created in 2002 and they would be 26-27 years old, i don't think its gona matter

Dude, I know, I was just adding my right answer to yours to balance out the first two incorrect answers in this thread. And so what if the thread is seven years old?

 

[maths-bbqz]

Dude, I know, I was just adding my right answer to yours to balance out the first two incorrect answers in this thread. And so what if the thread is seven years old?

So the guy's left school and finished uni and won't be checking BOS forums anymore, dumbass

 

[addikaye03]

Dude, I know, I was just adding my right answer to yours to balance out the first two incorrect answers in this thread. And so what if the thread is seven years old?

dude, i'm not snapping at you or anything, i was just saying that it doesnt matter that she got it wrong since they're finished school... but yes, you i and correct. Chill man

 

[scardizzle]

int. sin^2xcos^2x

int. 1/4(2sinxcosx)^2

1/4 int. sin^2 (2x)

[cos4x=1-2sin^2(2x)]

1/8 int. sin^2(2x)=1/8 int.

1/8(x+1/4sin4x)+C #

Isnt the intergral of (1-cos4x) = x-sin4x/4? And therefore wouldnt the answer be 1/8(x - sin4x/4) + C

Do correct me if im wrong

 

[edmund gosse]

dude, i'm not snapping at you or anything, i was just saying that it doesnt matter that she got it wrong since they're finished school... but yes, you i and correct. Chill man

I am chilled, my homey brother-man.

 

[edmund gosse]

So the guy's left school and finished uni and won't be checking BOS forums anymore

That's some razor-sharp thinking from you right there, maths-bbqz.

dumbass

Goodness! How unpleasant! Why don't you come over here and say that?

Oh, that's right. You can't, can you?

 

[jchoi]

Isnt the intergral of (1-cos4x) = x-sin4x/4? And therefore wouldnt the answer be 1/8(x - sin4x/4) + C

Do correct me if im wrong

It's correct. if y = sin x, dy/dx = cos x. This means Integral of sin x = cos x + C.
So, if we integrate (1-cos4x), it becomes [x-1/4(sin4x)] + C.

I think... lol it looks correct.