Original post by DryTowel
what q
Part b thanks
Original post by Anee_rk
Part b thanks
My clue to you is to substitute the denominator at the beginning. Try it by yourself. It is a bit tough, but you can do it.
(edited 1 year ago)
Original post by Kallisto
My clue to you is to substitute the denominator at the beginning. Try it by yourself. It is a bit tough, but you can do it.
using u=sqrt4-x^2 ?
Original post by Anee_rk
using u=sqrt4-x^2 ?
Yep, that is possible, but maybe too complicate for you in the next step to solve. You can do it without the square root too.
Original post by Anee_rk
Part b
Unless I'm missing an obvious trick, you can split this into 2 integrals - the first one should be a standard integral in your formula book and the second one can be tackled using the "obvious" substitution
This is the best method:
Substitute x = 2sinu
Root(4-x^2) = Root(4 - (2sinu)^2) = Root(4(cosu)^2).
So your denominator will be 2cosu.
dx = 2cosu du
So we have integral[ (2+x)/(2cosu) dx ]
= integral[ (2cosu)(2 + 2sinu)/(2cosu) du ]
= integral[ 2 + 2sinu du ]
= [ 2u - 2cosu ] between 0 and pi/6
= 2+ pi/3 - sqrt3
Substitute x = 2sinu
Root(4-x^2) = Root(4 - (2sinu)^2) = Root(4(cosu)^2).
So your denominator will be 2cosu.
dx = 2cosu du
So we have integral[ (2+x)/(2cosu) dx ]
= integral[ (2cosu)(2 + 2sinu)/(2cosu) du ]
= integral[ 2 + 2sinu du ]
= [ 2u - 2cosu ] between 0 and pi/6
= 2+ pi/3 - sqrt3
You can use integration by recognition…
Original post by DryTowel
This is the best method:...
Could you pls have a read of the forum guidelines about not posting solutions and pls delete. Thanks.
https://www.thestudentroom.co.uk/showthread.php?t=4919248
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