STEP Question integration by substitution Watch

Moordland
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http://www.maths.cam.ac.uk/undergrad...tep/advpcm.pdf

This is from page 4.

Use the substitution x = 2 − cos θ to evaluate the integral:


\displaystyle\int^2_0 (x-1) / (3-x)  dx

Where (x-1)/(3-x) is to the power of a half.


So far I have got integral with limits 2 to 1.5 of (1/cos(theta)/1+cos(theta))^0.5 sin(theta) dtheta
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Moordland
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I am stumped what do i do next pls
Any advice of what I should try and do
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username1763791
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You need to change that limits
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Renzhi10122
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(Original post by Moordland)
http://www.maths.cam.ac.uk/undergrad...tep/advpcm.pdf

This is from page 4.

Use the substitution x = 2 − cos θ to evaluate the integral:


\displaystyle\int^2_0 (x-1) / (3-x)  dx

Where (x-1)/(3-x) is to the power of a half.


So far I have got integral with limits 2 to 1.5 of (1/cos(theta)/1+cos(theta))^0.5 sin(theta) dtheta
multiply your fraction in the square root to get it in the form  \frac{a^2}{b^2}
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Moordland
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(Original post by Jordan\)
You need to change that limits
I have got the upper limit as 1 and the lower limit as 60! I am sure the upper limit is correct but the lower limit?

(Original post by TeeEm)
change the limitsmultiply top and bottom in the radical by (1-cosθ)

I can provide my working out if required.

Thanks.
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TeeEm
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(Original post by Moordland)
I am stumped what do i do next pls
Any advice of what I should try and do
change the limits
multiply top and bottom in the radical by (1-cosθ)
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CaptainJosh
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for the definite integral, indefinite is equal to 2 plus your constant
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TeeEm
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(Original post by Moordland)
I have got the upper limit as 1 and the lower limit as 60! I am sure the upper limit is correct but the lower limit?




I can provide my working out if required.

Thanks.
60!!!???
in Calculus there is no degrees...
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TeeEm
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(Original post by CaptainJosh)
for the definite integral, indefinite is equal to 2 plus your constant
are you sure?
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CaptainJosh
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(Original post by TeeEm)
are you sure?
Two in this case, rather x plus your constant.
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TeeEm
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(Original post by CaptainJosh)
Two in this case, rather x plus your constant.
I do not think so ...

I do not know what is your level but there might be a misconception here ...
The indefinite integral has a "vile" answer
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CaptainJosh
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(Original post by TeeEm)
I do not think so ...

I do not know what is your level but there might be a misconception here ...
The indefinite integral has a "vile" answer
I just finished my GCSEs, going into year 12 soon so I am certainly not the best
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TeeEm
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(Original post by CaptainJosh)
I just finished my GCSEs, going into year 12 soon so I am certainly not the best
fair enough ...
this is very hard Y13 work so a while to go.
All the best and keep learning.
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Moordland
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(Original post by TeeEm)
I do not think so ...

I do not know what is your level but there might be a misconception here ...
The indefinite integral has a "vile" answer
About the limits TeeEm:

So with 2 as the upper limit, there is:

x=2-cos(theta)
Substitute x=2
2=2-cos(theta)
0= -cos(theta)
cos(theta)=0
arccos(0) = theta = 1.

And for 1.5 as the lower lmit.

I get:
1.5=2-cos(theta)
-0.5=-cos(theta)
cos(theta)=0.5
cos^-1(0,5) = theta
theta=??
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TeeEm
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(Original post by Moordland)
About the limits TeeEm:

So with 2 as the upper limit, there is:

x=2-cos(theta)
Substitute x=2
2=2-cos(theta)
0= -cos(theta)
cos(theta)=0
arccos(0) = theta = 1.

And for 1.5 as the lower lmit.

I get:
1.5=2-cos(theta)
-0.5=-cos(theta)
cos(theta)=0.5
cos^-1(0,5) = theta
theta=??
cos(theta) = 0 implies theta is pi/2
cos(theta) = 1/2 implies theta is pi/3
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Moordland
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(Original post by TeeEm)
cos(theta) = 0 implies theta is pi/2
cos(theta) = 1/2 implies theta is pi/3
Thank you that seems nicer, may I ask why in radians is that by convention.
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TeeEm
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(Original post by Moordland)
Thank you that seems nicer, may I ask why in radians is that by convention.
All calculus is based in radians, otherwise you have a silly scaling factor.

You work in degrees only in 2 cases

Purely geometrical problems (mensurations including vector geometry but not vector calculus)
pure trigonometric equations if it indicates you are working in degrees.
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Moordland
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(Original post by TeeEm)
All calculus is based in radians, otherwise you have a silly scaling factor.

You work in degrees only in 2 cases

Purely geometrical problems (mensurations including vector geometry but not vector calculus)
pure trigonometric equations if it indicates you are working in degrees.
Great, thank you for the information.
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Moordland
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(Original post by TeeEm)
cos(theta) = 0 implies theta is pi/2
cos(theta) = 1/2 implies theta is pi/3
Sir, how do you do that without a calculator, I managed to do the first step, but not the second unfortunately.
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TeeEm
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(Original post by Moordland)
Sir, how do you do that without a calculator, I managed to do the first step, but not the second unfortunately.
just inverse cos in radians
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