2 Harder Integration with sin²x stuff. (1 Viewer)

[currysauce]

1. Find

INT cos^3 x dx


2. a) Find the area enclosed between y=sinx and y=cosx and the lines x=pi/6 , x=pi/4. I have done this.

ANSWER was 1/2 (2root(2) - root(3) -1)

b) This area is rotated about the x-axis. Find the volume of the solid of revolution formed.

ANSWER is pi/4(2-root(3))

THANKS

 

[mojako]

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

the derivative of sin x is cos x

let u = sin x
INT cos x * (1-sin^2 x) * dx
= INT du * (1-u^2)

2.
Vol
= pi*INT (cos x)^2 dx - pi*INT (sin x)^2 dx
= pi*INT (cos^2 x - sin^2 x) dx
= pi*INT (1 - 2sin^2 x) dx
= pi*INT (1 - 2[1/2 - 1/2 cos2x]) dx

hopefully its right.

 

[Slidey]

Integration by substitution without the sub provided is 4unit, mojako. The 3unit method used for that question is:

note that cos3@=4cos^3(@)-3cos@
.'. cos^3(@) = (cos3@+3cos@)/4
So integral of cos^3(@) = integral of (cos3@+3cos@)/4

 

[currysauce]

i can't seem to get the volume required by your method.

 

[Xayma]

Mojako:
cos²x -sin²x=cos 2x.

Makes the integral alot simplier.

 

[mojako]

Integration by substitution without the sub provided is 4unit, mojako. The 3unit method used for that question is:

note that cos3@=4cos^3(@)-3cos@
.'. cos^3(@) = (cos3@+3cos@)/4
So integral of cos^3(@) = integral of (cos3@+3cos@)/4

the expansion of cos3@ is rarely used in 3U as well...

i can't seem to get the volume required by your method.

pi times (integral of cos 2x dx from x=pi/6 to x=pi/4)
should give you
pi/4(2-root(3))

cos2x -sin2x=cos 2x.

That's right =p

 

[Slidey]

the expansion of cos3@ is rarely used in 3U as well...

It's still within the 3unit syllabus.

I'd be interested in a simpler method still within the scope of the 3unit syllabus, though.

 

[FinalFantasy]

INT cos^3 x dx
=int. cos²xd(sinx)
=int. (1-sin²x)d(sinx)
=sinx-sin³\3+c

is dat right? and in da 3unit scope?

 

[currysauce]

the expansion of cos3@ is rarely used in 3U as well...



pi times (integral of cos 2x dx from x=pi/6 to x=pi/4)
should give you
pi/4(2-root(3))


That's right =p

That doesn't work by my calculations.... besides i don't tihnk thats the question.

 

[mojako]

INT cos^3 x dx
=int. cos²xd(sinx)
=int. (1-sin²x)d(sinx)
=sinx-sin³\3+c

is dat right? and in da 3unit scope?

lol definitely not... its not even used in any hsc textbooks

this one is more like it:
INT cos^3 x dx
= INT cos x * cos^2 x dx
= INT cos x * (1-sin^2 x) dx
= 1 - 1/3 sin^3 x + C

hihihi

EDIT:
its sin x - 1/3 sin^3 x + C

 

[FinalFantasy]

lol definitely not... its not even used in any hsc textbooks

this one is more like it:
INT cos^3 x dx
= INT cos x * cos^2 x dx
= INT cos x * (1-sin^2 x) dx
= 1 - 1/3 sin^3 x + C

hihihi

how do u get the 1 in the "1 - 1/3 sin^3 x + C"

 

[mojako]

how do u get the 1 in the "1 - 1/3 sin^3 x + C"

I used Rikku's Mix ability to combine 2 rare healing items.

 

[FinalFantasy]

lol definitely not... its not even used in any hsc textbooks

this one is more like it:
INT cos^3 x dx
= INT cos x * cos^2 x dx
= INT cos x * (1-sin^2 x) dx
= 1 - 1/3 sin^3 x + C

hihihi

from ur "INT cos x * (1-sin^2 x) dx"
shouldn't u let u=sinx
du=cosxdx
so "INT cos x * (1-sin^2 x) dx"=int. (1-u²)du=u-u³\3=sinx-sin³\3+c

 

[currysauce]

guys.. how about the volume question

 

[Xayma]

Mojako's working should be right if you follow it through.

 

[currysauce]

Ya i got it!

Thanks all.