- #1
Haths
- 33
- 0
I have three questions, I don't need a full working through, but I'd prefer some hints or a simmilar example for where I am going wrong/need help. The answers arn't important, but the method of working them out is.
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A line AB is given;
Integ{ x dy - y dx }
Where the line is the peramitised set of equations: x=t2 and y=t+1 between 0<t<1
2. My attempt at a solution
I've assumed that this means that I am intergrating;
Integ{ t2 [d/dy] - t+1 [d/dx] } dt
Which is;
Integ{ -t2+2t } dt ==> 1/3 t3 + t2 |10
Therefore: 1.333...
I can't remember if this is what you do or not and considering I was going to ask some other questions I thought I'd ask this one too.
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Knowing the double angle formula for sine(a+b). Intergrate;
Integ{ sin(2x)cos(3x) } dx
My first thought was simply rearrage the double angle formula for the integrals;
Integ{ sin(5x) } dx - Integ{ cos(2x)sin(3x) } dx
But of course that gets me nowhere. Then I thought, what if I convert sin(2x) to 2sin(x)cos(x) and make my integral;
2*Integ{ sin(x)cos(x)cos(3x) } dx
But again that doesn't appear to help as I can't use a change of varible on the cos(3x) term, and if I want to expand that using the same double angle rules as before you get a nasty;
Integ{ 2sin(x)cos4(x) - 2sin3(x)cos2(x) - 4sin3(x)cos2(x) }
Which I suppose is doable, but there should be a trick for this problem, and that's what I'm missing
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Knowing d/du tan(u) = 1 + tan2(u)
Intergrate;
Integ{ 1 / (1 + x2) } dx
THEN find;
IntegInfinity0{ 1 / (3 + 2x2) }
For that last part I haven't a clue...
The first part however I assumed that I could let x = tan(u) and so change my integral to;
Integ{ 1 / tan(u) } du
Which using the product rule I believe intergrates to cos(u) Ln| sin(u) | + 1 as 1/tan = cos/sin
But I don't know is that's the right method as ignoring the 'hint' it gets nasty quickly :(.
Cheers and thanks in advance,
Haths
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Homework Statement
A line AB is given;
Integ{ x dy - y dx }
Where the line is the peramitised set of equations: x=t2 and y=t+1 between 0<t<1
2. My attempt at a solution
I've assumed that this means that I am intergrating;
Integ{ t2 [d/dy] - t+1 [d/dx] } dt
Which is;
Integ{ -t2+2t } dt ==> 1/3 t3 + t2 |10
Therefore: 1.333...
I can't remember if this is what you do or not and considering I was going to ask some other questions I thought I'd ask this one too.
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Homework Statement
Knowing the double angle formula for sine(a+b). Intergrate;
Integ{ sin(2x)cos(3x) } dx
The Attempt at a Solution
My first thought was simply rearrage the double angle formula for the integrals;
Integ{ sin(5x) } dx - Integ{ cos(2x)sin(3x) } dx
But of course that gets me nowhere. Then I thought, what if I convert sin(2x) to 2sin(x)cos(x) and make my integral;
2*Integ{ sin(x)cos(x)cos(3x) } dx
But again that doesn't appear to help as I can't use a change of varible on the cos(3x) term, and if I want to expand that using the same double angle rules as before you get a nasty;
Integ{ 2sin(x)cos4(x) - 2sin3(x)cos2(x) - 4sin3(x)cos2(x) }
Which I suppose is doable, but there should be a trick for this problem, and that's what I'm missing
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Homework Statement
Knowing d/du tan(u) = 1 + tan2(u)
Intergrate;
Integ{ 1 / (1 + x2) } dx
THEN find;
IntegInfinity0{ 1 / (3 + 2x2) }
The Attempt at a Solution
For that last part I haven't a clue...
The first part however I assumed that I could let x = tan(u) and so change my integral to;
Integ{ 1 / tan(u) } du
Which using the product rule I believe intergrates to cos(u) Ln| sin(u) | + 1 as 1/tan = cos/sin
But I don't know is that's the right method as ignoring the 'hint' it gets nasty quickly :(.
Cheers and thanks in advance,
Haths