# Integration

• May 5th 2007, 04:07 PM
r_maths
Integration
• May 5th 2007, 04:10 PM
Jhevon
Quote:
you are correct up to the third to last line, you expanded incorrectly. the answer you should get is the one you said you should

(1/2)(pi/4 - sin(2*(pi/4))/2) = pi/8 - (1/2)(1/2) = pi/8 - 1/4

try the second question now
• May 5th 2007, 04:18 PM
r_maths
Where did you get the bit in bold?
(1/2)(pi/4 - sin(2*(pi/4))/2) = pi/8 - (1/2)(1/2) = pi/8 - 1/4

http://img246.imageshack.us/img246/2208/int003ia1.png
From that, it gives 0. :S
• May 5th 2007, 04:21 PM
Jhevon
Quote:

Originally Posted by r_maths
Where did you get the bit in bold?
(1/2)(pi/4 - sin(2*(pi/4))/2) = pi/8 - (1/2)(1/2) = pi/8 - 1/4

http://img246.imageshack.us/img246/2208/int003ia1.png
From that, it gives 0. :S

i only did the first piece, the second function is zero so i forgot about it.

sin[2*(pi/4)] = sin(pi/2) = 1

so, (1/2)(pi/4 - sin(2*(pi/4))/2) = (1/2)(pi/4 - 1/2) = (1/2)(pi/4) - (1/2)(1/2)
• May 5th 2007, 04:25 PM
r_maths
O thanks alot. I can't believe I didn't see that.
Im tired and I need sleep.
Thanks :)
• May 7th 2007, 12:47 PM
r_maths
Sorry, have to bump this thread again.
Its question number 2.

The full question:
http://img256.imageshack.us/img256/3835/int004ke6.png

I've done part a.

http://img49.imageshack.us/img49/2749/int005tn7.png

Thanks
• May 7th 2007, 05:58 PM
Jhevon
Quote:

Originally Posted by r_maths
Sorry, have to bump this thread again.
Its question number 2.

The full question:
http://img256.imageshack.us/img256/3835/int004ke6.png

I've done part a.

http://img49.imageshack.us/img49/2749/int005tn7.png

Thanks

ok, so the first question made no sense. it told you to find the answer by writing the answer...but anyway.

for the second question, i would not use that. i would say that:

cos^3x = cos^2x*cosx = (1 - sin^2x)cosx

then use the substitution u = sinx, we'd end up integrating 1 - u^2 ...easy. but in the spirit of being as weird as this question is, let's do it like they said we should.

we are told that

cos(3x) = 4cos^3x - 3cosx
=> 4cos^3x = cos(3x) + 3cosx
=> cos^3x = (1/4)[cos(3x) + 3cosx]
=> int{cos^3x}dx = (1/4)*int{[cos(3x) + 3cosx]}dx
you should be able to find the integral on the right without much problems...if you have problems, don't be afraid to say so
• May 7th 2007, 06:01 PM
topsquark
Quote:

Originally Posted by Jhevon
for the second question, i would not use that. i would say that:

cos^3x = cos^2x*cosx = (1 - sin^2x)cosx

The problem is dealing with cos(3x) not cos^3(x).

For the record:
cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)

= (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)

= 2cos^3(x) - cos(x) - 2sin^2(x)cos(x)

= 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)

= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)

= 4cos^3(x) - 3cos(x)

-Dan
• May 7th 2007, 06:03 PM
Jhevon
Quote:

Originally Posted by topsquark
The problem is dealing with cos(3x) not cos^3(x).

-Dan

what do you mean? (b) says find int{cos^3x}dx
• May 7th 2007, 06:05 PM
topsquark
Quote:

Originally Posted by Jhevon
what do you mean? (b) says find int{cos^3x}dx

Ah. :o I didn't look at that part.

I think they want to use the identity as
cos^3(x) = (1/4)cos(3x) + (3/4)cos(x)

which is easier to integrate.

-Dan
• May 7th 2007, 06:12 PM
Jhevon
Quote:

Originally Posted by topsquark
Ah. :o I didn't look at that part.

I think they want to use the identity as
cos^3(x) = (1/4)cos(3x) + (3/4)cos(x)

which is easier to integrate.

-Dan

which is what i said.

and in regards to your second earlier post, this is what the question says.

let the formula they have for cos(3x) be f(x)

the question basically said.

by writing cos(3x) as cos(3x) = f(x) show that cos(3x) = f(x)

if that's not redundant i don't know what is
• May 8th 2007, 04:19 AM
r_maths
Quote:

Originally Posted by Jhevon
ok, so the first question made no sense. it told you to find the answer by writing the answer...but anyway.

It was one of the proof ones were you had to show/prove it just like what topsquark did...
Quote:

cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)

= (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)

= 2cos^3(x) - cos(x) - 2sin^2(x)cos(x)

= 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)

= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)

= 4cos^3(x) - 3cos(x)
• May 8th 2007, 06:10 AM
Jhevon
Quote:

Originally Posted by r_maths
It was one of the proof ones were you had to show/prove it just like what topsquark did...

yeah, i know what topsquark did was correct, i'm just saying that's not how the question was phrased. it should have said something like:

"Prove that cos(3x) = 4cos^3x - 3cosx"

and that's it.

but since no one is seeing my point but me, let's just drop it. you have the correct answers:D
• May 8th 2007, 06:13 AM
topsquark
Quote:

Originally Posted by Jhevon
yeah, i know what topsquark did was correct, i'm just saying that's not how the question was phrased. it should have said something like:

"Prove that cos(3x) = 4cos^3x - 3cosx"

and that's it.

but since no one is seeing my point but me, let's just drop it. you have the correct answers:D

I agree: the first part of the question, at least, was badly worded.

-Dan