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Collection of P3 integration Questions
These are the questions I got stuck on today, I will rep everyone who answers questions correctly, but please be patient if there are many different people answering as I can only give 1 rep per 24 hrs 
1) integrate: x/sqrt(x-1)
2) integrate: cos2xsinx
3) integrate: sinxcos(1/2)x
4) An ellipse is given by:
x=3cost , y=sint 0<t<2pi
a) find the area of the finite region bounded by the epllipse and the positive x and y-axis
b) the region is rotated completely about the y-axis to form a solid of revolustion. find the volume of this solid.
Many thanks for any solutions
1) integrate: x/sqrt(x-1)
2) integrate: cos2xsinx
3) integrate: sinxcos(1/2)x
4) An ellipse is given by:
x=3cost , y=sint 0<t<2pi
a) find the area of the finite region bounded by the epllipse and the positive x and y-axis
b) the region is rotated completely about the y-axis to form a solid of revolustion. find the volume of this solid.
Many thanks for any solutions
For cos2x.sinx...
Use Integration by parts twice to get:
INT cos2xsinx = -cosxcos2x - 2INTsin2xcosx
Another integration by parts,
INTsin2xcos = sin2xsinx - 2INTcos2xsinx
So that
INTcos2xsinx = -cosxcos2x - 2(sin2xsinx - 2INTcos2xsinx)
INTcos2xsinx = -cosxcos2x - 2sin2xsinx + 4INTcos2xsinx
INTcos2xsinx - 4INTcos2xsinx = -cosxcos2x - 2sin2xsinx
--> -3INTcos2xsinx = -cosxcos2x - 2sin2xsinx
INTcos2xsinx = (-1/3)(-cosxcos2x - 2sin2xsinx)
-->INTcos2xsinx = (1/3)(2sin2xsinx + cos2xcosx) + C
Use Integration by parts twice to get:
INT cos2xsinx = -cosxcos2x - 2INTsin2xcosx
Another integration by parts,
INTsin2xcos = sin2xsinx - 2INTcos2xsinx
So that
INTcos2xsinx = -cosxcos2x - 2(sin2xsinx - 2INTcos2xsinx)
INTcos2xsinx = -cosxcos2x - 2sin2xsinx + 4INTcos2xsinx
INTcos2xsinx - 4INTcos2xsinx = -cosxcos2x - 2sin2xsinx
--> -3INTcos2xsinx = -cosxcos2x - 2sin2xsinx
INTcos2xsinx = (-1/3)(-cosxcos2x - 2sin2xsinx)
-->INTcos2xsinx = (1/3)(2sin2xsinx + cos2xcosx) + C
Freddy C
3) integrate: sinxcos(1/2)x
Cos(x/2)=+-(1/rt2).rt(cosx+1). Plugging in you want to integrate,
+-(1/rt(2))INT(sinx.rt(cosx + 1)). Let u=cosx+1. Then du/dx=-sinx.
--> dx=du/-sinx
-->
+-(1/rt(2))INT(-rt(u)) --> -+(1/rt(2))INT(u^1/2) = -+(1/rt(2)).u^(3/2).2/3
= -+(rt(2)/3).(cosx+1)^(3/2) + C
4) An ellipse is given by:
x=3cost , y=sint 0<t<2pi
a) find the area of the finite region bounded by the epllipse and the positive x and y-axis
3pi/4 (because the area of the ellipse x^2/a^2+y^2/b^2=1 is pi a b)
b) the region is rotated completely about the y-axis to form a solid of revolustion. find the volume of this solid.
12pi (because the volume of the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2=1 is 4 pi a b c/3)
x=3cost , y=sint 0<t<2pi
a) find the area of the finite region bounded by the epllipse and the positive x and y-axis
3pi/4 (because the area of the ellipse x^2/a^2+y^2/b^2=1 is pi a b)
b) the region is rotated completely about the y-axis to form a solid of revolustion. find the volume of this solid.
12pi (because the volume of the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2=1 is 4 pi a b c/3)
C4>O7
1) Using the substitution u²=x-1 =>x=u²+1, dx/du=2u =>dx=2u.du
∫(u²-1)/u.dx = ∫(u²-1)/u.2u.du = 2∫(u²-1).du = 2(1/3u³-u)+C = 2/3(x-1)^(³/2) -2√(x-1)+C
∫(u²-1)/u.dx = ∫(u²-1)/u.2u.du = 2∫(u²-1).du = 2(1/3u³-u)+C = 2/3(x-1)^(³/2) -2√(x-1)+C
Thanks for the answer: the book says:
(2/3)(x+2)√(x-1)
which confused me
I know you can take √(x-1) as the common factor, but I don't see how they got the (x+2)
I get (2/3)(x+4)√(x-1)
i used the same method, but had a different answer for q1.
Someone find the blindingly stupid mistake ¬_¬
∫x √(x-1)
let u = x-1
du = dx
=> ∫(u + 1).√u
= ∫u^1.5 + √u
=(1/2.5)u^2.5 + (1/1.5)u^1.5
subbing x back in.
= 1/2.5(x-1)^2.5 + 1/1.5(x-1)^1.5
Someone find the blindingly stupid mistake ¬_¬
∫x √(x-1)
let u = x-1
du = dx
=> ∫(u + 1).√u
= ∫u^1.5 + √u
=(1/2.5)u^2.5 + (1/1.5)u^1.5
subbing x back in.
= 1/2.5(x-1)^2.5 + 1/1.5(x-1)^1.5
totally different method for question 2 too.
∫cos2xsinx dx
but cos2x = 1-2sin^2 X
=> ∫(1-2sin^2x).sinx dx
=> ∫ sinx - 2sin^3x dx
let u = sin x
du = cosx dx
∫u - 2u^3 du
= cosx - 2/4u^4. 1/cosx
= cosx - 1/2sin^4x /cosx
∫cos2xsinx dx
but cos2x = 1-2sin^2 X
=> ∫(1-2sin^2x).sinx dx
=> ∫ sinx - 2sin^3x dx
let u = sin x
du = cosx dx
∫u - 2u^3 du
= cosx - 2/4u^4. 1/cosx
= cosx - 1/2sin^4x /cosx
Freddy:
C4>O7 made a small slip: the nominator should've been u²+1 and not u²-1. That should give you:
(2/3)(x-1)^(3/2) + 2(x-1)^(1/2)
Taking (x-1)^(1/2) as a common factor would then get you the answer you need.
nicholls2k:
For the first one: it's x/sqrt(x-1), not x * sqrt(x-1).
For the second one: you're not using substitutions properly. You said du = cosx dx, but there's no cosx you can substitute, so you can't just simply replace dx by du. And I'm not really sure how you got the last two lines.
C4>O7 made a small slip: the nominator should've been u²+1 and not u²-1. That should give you:
(2/3)(x-1)^(3/2) + 2(x-1)^(1/2)
Taking (x-1)^(1/2) as a common factor would then get you the answer you need.
nicholls2k:
For the first one: it's x/sqrt(x-1), not x * sqrt(x-1).
For the second one: you're not using substitutions properly. You said du = cosx dx, but there's no cosx you can substitute, so you can't just simply replace dx by du. And I'm not really sure how you got the last two lines.
J.F.N
Cos(x/2)=+-(1/rt2).rt(cosx+1). Plugging in you want to integrate,
+-(1/rt(2))INT(sinx.rt(cosx + 1)). Let u=cosx+1. Then du/dx=-sinx.
--> dx=du/-sinx
-->
+-(1/rt(2))INT(-rt(u)) --> -+(1/rt(2))INT(u^1/2) = -+(1/rt(2)).u^(3/2).2/3
= -+(rt(2)/3).(cosx+1)^(3/2) + C
+-(1/rt(2))INT(sinx.rt(cosx + 1)). Let u=cosx+1. Then du/dx=-sinx.
--> dx=du/-sinx
-->
+-(1/rt(2))INT(-rt(u)) --> -+(1/rt(2))INT(u^1/2) = -+(1/rt(2)).u^(3/2).2/3
= -+(rt(2)/3).(cosx+1)^(3/2) + C
The answer in the book is (-1/3)(cos(3x/2))-(cos(x/2))
any ideas how they got to that?
Thanks for the answers
Using the identity:
sinA + sinB = 2sin(A+B/2)cos(A-B/2)
with A=(3/2)x and B=(1/2)x, we get:
sinxcos(x/2) = (sin(3x/2) + sin(x/2))/2
So, the integral becomes:
I = (1/2) ∫ sin(3x/2) dx + (1/2) ∫ sin(x/2) dx
= (1/2) . (2/3) . -cos(3x/2) + (1/2) . 2 . -cos(x/2) + C
= (-1/3) cos(3x/2) - cos(x/2) + C
Another way you can do the integral is by using the identity:
sinx = 2sin(x/2)cos(x/2)
Then the integral becomes:
I = ∫ 2sin(x/2)cos²(x/2) dx
= -4 ∫ -(1/2)sin(x/2).cos²(x/2) dx
Using the substitution u=cos(x/2) with du=-(1/2)sin(x/2)dx, we get:
I = -4 ∫ u² du
= (-4/3)u³ + C
= (-4/3)[cos(x/2)]³ + C
You should note that all the answers are correct (as you can verify by differentiating); they're just a constant apart.
sinA + sinB = 2sin(A+B/2)cos(A-B/2)
with A=(3/2)x and B=(1/2)x, we get:
sinxcos(x/2) = (sin(3x/2) + sin(x/2))/2
So, the integral becomes:
I = (1/2) ∫ sin(3x/2) dx + (1/2) ∫ sin(x/2) dx
= (1/2) . (2/3) . -cos(3x/2) + (1/2) . 2 . -cos(x/2) + C
= (-1/3) cos(3x/2) - cos(x/2) + C
Another way you can do the integral is by using the identity:
sinx = 2sin(x/2)cos(x/2)
Then the integral becomes:
I = ∫ 2sin(x/2)cos²(x/2) dx
= -4 ∫ -(1/2)sin(x/2).cos²(x/2) dx
Using the substitution u=cos(x/2) with du=-(1/2)sin(x/2)dx, we get:
I = -4 ∫ u² du
= (-4/3)u³ + C
= (-4/3)[cos(x/2)]³ + C
You should note that all the answers are correct (as you can verify by differentiating); they're just a constant apart.
dvs
nicholls2k:
For the first one: it's x/sqrt(x-1), not x * sqrt(x-1).
.
nicholls2k:
For the first one: it's x/sqrt(x-1), not x * sqrt(x-1).
.
good point well presented, back to the drawing board :<
dvs
For the second one: you're not using substitutions properly. You said du = cosx dx, but there's no cosx you can substitute, so you can't just simply replace dx by du. And I'm not really sure how you got the last two lines.
For the second one: you're not using substitutions properly. You said du = cosx dx, but there's no cosx you can substitute, so you can't just simply replace dx by du. And I'm not really sure how you got the last two lines.
yes that would make sence, is there any way i can integrate
sinx - 2(sin^3)x
or would i have to use the "other" method?
thanks in advace
∫ sinx - 2sin³x dx = ∫ sinx dx - ∫ 2sin³x dx
= -cosx - 2 ∫ sin³x dx
So the tricky part is integrating sin³x. There are a couple of ways you can go about this:
1. Manipulation, trig identities and a substitution (which is, in my opinion, the better method);
2. The "other" method.
(Or you could just use a reduction formula, but that's not introduced until P5...)
I = ∫ sin³x dx
= ∫ sinx sin²x dx
= ∫ sinx (1 - cos²x) dx
= ∫ sinx dx + ∫ -sinx cos²x dx
To do the second integral, use the substitution u=cosx with du = -sinx dx. Then:
I = -cosx + ∫ u² du
= -cosx + (1/3)u³ + C
= -cosx + (1/3)cos³x + C
= -cosx - 2 ∫ sin³x dx
So the tricky part is integrating sin³x. There are a couple of ways you can go about this:
1. Manipulation, trig identities and a substitution (which is, in my opinion, the better method);
2. The "other" method.
(Or you could just use a reduction formula, but that's not introduced until P5...)
I = ∫ sin³x dx
= ∫ sinx sin²x dx
= ∫ sinx (1 - cos²x) dx
= ∫ sinx dx + ∫ -sinx cos²x dx
To do the second integral, use the substitution u=cosx with du = -sinx dx. Then:
I = -cosx + ∫ u² du
= -cosx + (1/3)u³ + C
= -cosx + (1/3)cos³x + C
1.
BC . AC = |BC| |AC| cos(BCA)
BC = (1-2, 5-0, -1-5) = (-1, 5, -6)
AC = (1+1, 5-2, -1+1) = (2, 3, 0)
|BC| = sqrt(62)
|AC| = sqrt(13)
cos(BCA) = (BC . CA)/(|BC| |CA|) = (-2 + 15)/sqrt(62*13) = 13/sqrt(806)
Area of ABC:
0.5 * |BC| * |AC| * sin(BCA) = 0.5 * sqrt(806) * sqrt[1-cos(BCA)²] = 0.5 sqrt(806) sqrt(49/62) = 0.5 sqrt(637)
2.
Draw a triangle with the given information. (See attachment.)
Using the cosine rule:
|a-b|² = |a|² + |b|² - 2 |a| |b| cosθ
49 = 9 + 25 - 30cosθ
cosθ = -0.5
θ = 2pi/3
BC . AC = |BC| |AC| cos(BCA)
BC = (1-2, 5-0, -1-5) = (-1, 5, -6)
AC = (1+1, 5-2, -1+1) = (2, 3, 0)
|BC| = sqrt(62)
|AC| = sqrt(13)
cos(BCA) = (BC . CA)/(|BC| |CA|) = (-2 + 15)/sqrt(62*13) = 13/sqrt(806)
Area of ABC:
0.5 * |BC| * |AC| * sin(BCA) = 0.5 * sqrt(806) * sqrt[1-cos(BCA)²] = 0.5 sqrt(806) sqrt(49/62) = 0.5 sqrt(637)
2.
Draw a triangle with the given information. (See attachment.)
Using the cosine rule:
|a-b|² = |a|² + |b|² - 2 |a| |b| cosθ
49 = 9 + 25 - 30cosθ
cosθ = -0.5
θ = 2pi/3
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