# Further Integration C4Watch

#1
I have tried these questions, but couldn't do them. I will post my attempt, and the question.

1) INT: x/(2e^x) dx. Therefore INT: (xe^-x)/2 dx

INT by parts : u=x/2 u'=1/2 v'=(e^-x)/2 v=(-e^-x)/2

Therefore, (-xe^-x)/4 - INT (-e^-x)/4 dx

So, (-xe^-x)/4 - (e^-x)/4 + c

I'm not sure what i have done wrong.

2) INT: xcosec^2.x dx

u=x u'=1 v'=cosec^2.x v'=1/(1-cos^2.x)= 1-sec^2.x , so v=x-tanx

Therefore x^2 - xtanx - INT 1(x-tanx) dx
=x^2 - tanx - (x^2)/2 + ln(secx) +c

Again I'm not sure what i have done wrong.

3) FInd the volume (rotated 360 around x axis) under the curve 2xe^x with limits 0 and -1.

pi INT y^2 dx
pi INT 4x^2.e^2x dx

u=4x^2 u' =8x v'=e^2x v=1/2(e^2x)

So, pi[2x^2.e^2x] - pi INT 4x.e^2x dx
u=4x u'=4 v'=e^2x v=1/2(e^2x)

Therefore: pi[2x^2.e^2x - 2x.e^2x -e^2x] Limits: x=0, x=-1
So, pi[(-e^0)-(2e^-2 + 2e^-2 - e^-2)
= pi[-1 - 3e^-2] = pi[3e^-2 - 1]

Thats the wrong answer.. but again i don't see why.

4) INT: ln(x-1)^1/2 dx

u=ln(x-1)^1/2 u'=1/2(x-1)^-1 v'=1 v=x

So, xln(x-1)^1/2 - INT x/2(x-1)^-1

=xln(x-1)^1/2 - (x/2)/(x-1)

I'm stuck on this question from here.

5)INT: e^3x.cos2x dx

u=e^3x u'=3e^3x v'=cos2x v=1/2(sin2x)

e^3x.sin2x - INT (3e^3x.sin2x)/2 dx

u=(3e^3x)/2 u'=(9e^3x)/2 v'=sin2x/2 v=-cos2x/4

e^3x.sin2x - (3e^3x.cos2x)/8 + INT (9e^3x.cos2x)/8 dx

I'm digging myself into a hole!

I appreciate any help. Thanks
0
12 years ago
#2
1) You've included the 1/2 twice. Once in u and once in v'. If you find uv', you get x/(4e^x). So take u=x/2 and v'=e^-x OR u=x and v'=(e^-x)/2.
0
12 years ago
#3
(Original post by Srathmore)

2) INT: xcosec^2.x dx

[I]u=x u'=1 v'=cosec^2.x v'=1/(1-cos^2.x)= 1-sec^2.x , so v=x-tanx
v'=1/(1-cos^2.x) is not the same as 1-sec^2.x
To see this think about 1/(1-x). This isn't the same as 1 - 1/x

Note that the derivative of cot(x) is -cosec^2.x. Hence v=-cotx
0
12 years ago
#4
(Original post by Srathmore[B)
5)INT: e^3x.cos2x dx[/B]

u=e^3x u'=3e^3x v'=cos2x v=1/2(sin2x)

e^3x.sin2x - INT (3e^3x.sin2x)/2 dx

u=(3e^3x)/2 u'=(9e^3x)/2 v'=sin2x/2 v=-cos2x/4

e^3x.sin2x - (3e^3x.cos2x)/8 + INT (9e^3x.cos2x)/8 dx

I'm digging myself into a hole!

I appreciate any help. Thanks
Let I=INT e^3x.cos2x dx

As you've said, u=e^3x, u'= 3e^3x, v'=cos2x v=(1/2)sin2x

So I = (1/2)e^3x.sin2x - (3/2) INT e^3x.sin2x dx

u= e^3x, u'=3e^3x, v'=-sin2x, v=(1/2)cos2x

So I = (1/2)e^3x.sin2x + (3/2)[ (1/2)e^3x.cos2x - (3/2) INT e^3xcos2x dx]

Notice that the last integral is just a scale factor of I
So I = (1/2)e^3x.sin2x + (3/4)e^3x.cos2x - (9/4) I

So (13/4) I = (1/2)e^3x.sin2x + (3/4)e^3x.cos2x

So I = INT e^3x.cos2x dx = (2/13)e^3x.sin2x + (4/13)e^3x.cos2x

Hopefully I haven't made any arithmetical errors.
0
#5
(Original post by mikeyT)
Notice that the last integral is just a scale factor of I
So I = (1/2)e^3x.sin2x + (3/4)e^3x.cos2x - (9/4) I
Where does the -9/4 come from?
0
12 years ago
#6
(Original post by Srathmore)
Where does the -9/4 come from?
(3/2)*(-3/2)
0
#7
(Original post by mikeyT)
(3/2)*(-3/2)
What i meant was why do you use -9/4. Surely, you would just take it over to the other side, and get 5/2 INT........
0
12 years ago
#8
where do you get 5/2 from?
0
#9
(Original post by mikeyT)
where do you get 5/2 from?
1+3/2.

Because you have the integral + 3/2 of the integral.
0
12 years ago
#10
If you read it carefully it says

(3/2)[ (1/2)e^3x.cos2x - (3/2) INT e^3xcos2x dx]

Notice the 3/2 outside the [....] and the 3/2 multiplying the integral.
0
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