A Level further maths- Integration using trig functions Watch

Yodalam
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Hi,

So I got this integral I can't seem to integrate:

Integral(limits 1.5 , 0.5) 1/(4x^2 -4x+5)
So I first completed the square (correctly)
Then I integrated it:
(Answer says there is a 0.5 here)[0.5arctan((2x-1)/2)] with limits 1.5 and 0.5
But this is somehow wrong because there is supposedly a factor of 0.5. Does anyone know where this extra 0.5 come from?

Thanks
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the bear
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(Original post by Yodalam)
Hi,

So I got this integral I can't seem to integrate:

Integral(limits 1.5 , 0.5) 1/(4x^2 -4x+5)
So I first completed the square (correctly)
Then I integrated it:
(Answer says there is a 0.5 here)[0.5arctan((2x-1)/2)] with limits 1.5 and 0.5
But this is somehow wrong because there is supposedly a factor of 0.5. Does anyone know where this extra 0.5 come from?

Thanks
i think they made a mistake.... it should be 0.25 in front, not 0.5
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Yodalam
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(Original post by the bear)
i think they made a mistake.... it should be 0.25 in front, not 0.5
The constant at the front is 0.25, I got 0.5. So there is an extra 0.5 and I am confused about where that extra number comes from
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the bear
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(Original post by Yodalam)
The constant at the front is 0.25, I got 0.5. So there is an extra 0.5 and I am confused about where that extra number comes from
are you saying the book agrees with me ?

:beard:
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RDKGames
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(Original post by Yodalam)
Hi,

So I got this integral I can't seem to integrate:

Integral(limits 1.5 , 0.5) 1/(4x^2 -4x+5)
So I first completed the square (correctly)
Then I integrated it:
(Answer says there is a 0.5 here)[0.5arctan((2x-1)/2)] with limits 1.5 and 0.5
But this is somehow wrong because there is supposedly a factor of 0.5. Does anyone know where this extra 0.5 come from?

Thanks
It's because 4x^2 - 4x + 5 = (2x-1)^2 + 2^2 therefore we have


\displaystyle \int \dfrac{dx}{(2x-1)^2 + 2^2}

Now let u = 2x-1 then dx = \dfrac{1}{2} du and we obtain


\displaystyle \dfrac{1}{2} \int \dfrac{du}{u^2 + 2^2}

and obviously this evaluates to \dfrac{1}{2} \cdot \dfrac{1}{2} \arctan \dfrac{u}{2} between the limits you got.
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Yodalam
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(Original post by the bear)
are you saying the book agrees with me ?

:beard:
Yes
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the bear
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(Original post by Yodalam)
Yes
Name:  tsrtriggy.png
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.... it should say (2x-1)/2 in the last line
Last edited by the bear; 1 month ago
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Yodalam
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(Original post by RDKGames)
It's because 4x^2 - 4x + 5 = (2x-1)^2 + 2^2 therefore we have


\displaystyle \int \dfrac{dx}{(2x-1)^2 + 2^2}

Now let u = 2x-1 then dx = \dfrac{1}{2} du and we obtain


\displaystyle \dfrac{1}{2} \int \dfrac{du}{u^2 + 2^2}

and obviously this evaluates to \dfrac{1}{2} \cdot \dfrac{1}{2} \arctan \dfrac{u}{2} between the limits you got.
Thanks,

So do I have to use substitution whenever there is a constant next to x in the integral? Because normally when there isnt an a constant next to x, substitiution is not needed. (I am talking about the (2x-1)^2 the 2 next to the x inside the bracket btw.
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the bear
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(Original post by Yodalam)
Thanks,

So do I have to use substitution whenever there is a constant next to x in the integral? Because normally when there isnt an a constant next to x, substitiution is not needed. (I am talking about the (2x-1)^2 the 2 next to the x inside the bracket btw.
the standard result refers to the integral of 1/( x2 + a2 )
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RDKGames
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(Original post by Yodalam)
Thanks,

So do I have to use substitution whenever there is a constant next to x in the integral? Because normally when there isnt an a constant next to x, substitiution is not needed. (I am talking about the (2x-1)^2 the 2 next to the x inside the bracket btw.
You should be aiming to use a substitution all the time whenever the integral isn't in its standard form. The very point of this is reduce it *to* a standard form from which you can just smack on the known results.
Last edited by RDKGames; 1 month ago
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Yodalam
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Ok. thanks for your help!
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Vinny C
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Np... is fun and practice. We used to attempt these in our heads when I was 20. No solution implies there is none... but you never know. You may come up with the warp field!
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Vinny C
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Maxwell... Newton, all the greats came up with tricks no-one had previously considered. Maths is like magic... voila... a rabbit! WTF? Show me again!
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