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A Level further maths- Integration using trig functions

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
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
Reply 2
Avatar for Yodalam
Yodalam
OP
14
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
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:
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 4x2−4x+5=(2x−1)2+224x^2 - 4x + 5 = (2x-1)^2 + 2^2 therefore we have


∫dx(2x−1)2+22\displaystyle \int \dfrac{dx}{(2x-1)^2 + 2^2}

Now let u=2x−1u = 2x-1 then dx=12dudx = \dfrac{1}{2} du and we obtain


12∫duu2+22\displaystyle \dfrac{1}{2} \int \dfrac{du}{u^2 + 2^2}

and obviously this evaluates to 12⋅12arctan⁡u2\dfrac{1}{2} \cdot \dfrac{1}{2} \arctan \dfrac{u}{2} between the limits you got.
Reply 5
Avatar for Yodalam
Yodalam
OP
14
Original post by the bear
are you saying the book agrees with me ?

:beard:

Yes
Original post by Yodalam
Yes

tsrtriggy.png

.... it should say (2x-1)/2 in the last line :colondollar:
(edited 4 years ago)
Reply 7
Avatar for Yodalam
Yodalam
OP
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Original post by RDKGames
It's because 4x2−4x+5=(2x−1)2+224x^2 - 4x + 5 = (2x-1)^2 + 2^2 therefore we have


∫dx(2x−1)2+22\displaystyle \int \dfrac{dx}{(2x-1)^2 + 2^2}

Now let u=2x−1u = 2x-1 then dx=12dudx = \dfrac{1}{2} du and we obtain


12∫duu2+22\displaystyle \dfrac{1}{2} \int \dfrac{du}{u^2 + 2^2}

and obviously this evaluates to 12⋅12arctan⁡u2\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.
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 )
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.
(edited 4 years ago)
Reply 10
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Yodalam
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Ok. thanks for your help!
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!
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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