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Trig Substitution for Integration

x(3−4x−4x2)12 \frac{x}{(3-4x-4x^2)^{\frac{1}{2}}}

The limits of integration on this are from -0.5 to 0.

I completed the square and used the Trig Substitution 2x+1 = 2SinQ


At the end after integrationI was left with −CosQ2−Q4 \frac{-CosQ}{2} - \frac{Q}{4}

Now I dont particularly like changing everything back into x just to evaluate the integral so I changed the limits to suit my substitution.

Basically when x = -0.5 , Q = 0
when x = 0 Q = pi/6

However my final answer is not right according to the book. It is right however if I substitute all the way back and use the original x values. So my question is, what is wrong with my new limits of integration?
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Avatar for davros
davros
16
Original post by Zilch
x(3−4x−4x2)12 \frac{x}{(3-4x-4x^2)^{\frac{1}{2}}}

The limits of integration on this are from -0.5 to 0.

I completed the square and used the Trig Substitution 2x+1 = 2SinQ


At the end after integrationI was left with −CosQ2−Q4 \frac{-CosQ}{2} - \frac{Q}{4}

Now I dont particularly like changing everything back into x just to evaluate the integral so I changed the limits to suit my substitution.

Basically when x = -0.5 , Q = 0
when x = 0 Q = pi/6

However my final answer is not right according to the book. It is right however if I substitute all the way back and use the original x values. So my question is, what is wrong with my new limits of integration?


I haven't checked your integration but you should get exactly the same answer if you've substituted the limits correctly. Is your answer vastly different, or just a negative sign out, or something else?
Reply 2
Avatar for Zilch
Zilch
OP
1
Original post by davros
I haven't checked your integration but you should get exactly the same answer if you've substituted the limits correctly. Is your answer vastly different, or just a negative sign out, or something else?


I did it myself at first and thought it was an algebra error but then I did it online on wolfgram and noticed the same difference.

The answer I'm getting is roughly 0.032 and the actual answer is -0.06
(edited 11 years ago)
Reply 3
Avatar for davros
davros
16
Original post by Zilch
I did it myself at first and thought it was an algebra error but then I did it online on wolfgram and noticed the same difference.

The answer I'm getting is roughly 0.032 and the actual answer is -0.06


Well I just put your Q limits into your Q function and got -0.06391 approximately so the error is in your arithemtic. I ended up evaluating:

(-0.5cos(pi/6) - (pi/24)) - (-0.5 - 0)

Note that your original function is negative throughout the range of integration, so you know that the answer has to be negative!
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Avatar for Zilch
Zilch
OP
1
Original post by davros
Well I just put your Q limits into your Q function and got -0.06391 approximately so the error is in your arithemtic. I ended up evaluating:

(-0.5cos(pi/6) - (pi/24)) - (-0.5 - 0)

Note that your original function is negative throughout the range of integration, so you know that the answer has to be negative!


Right, thanks. Made a slight mistake in the end where I integrated x/4 instead of 1/4. That made it x^2/8 in the online calc.

Thanks for the help!

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