q. Integrate sinxcosx/(cos2x+3)^1/2
i got sin2x/2*(cos2x+3)^-1/2
then -cos2x/4*(cos2x+3)^-1/2
then -(cos2x+3)^1/2 /4*0.5*2
hence I GOT, -(cos2x+3)^1/2 /4
but the answer is -(cos2x+3)^1/2 /2 :/
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SHUAIB_AKA_SHU
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- 04-04-2011 19:32
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SHUAIB_AKA_SHU
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- 04-04-2011 19:36
.......
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soutioirsim
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- 04-04-2011 19:44
Did you remember to divide by
?
(If you use a u-sub of "u = cos2x" you end up having to divide by, you could of missed it out...)
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SHUAIB_AKA_SHU
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- 04-04-2011 19:45
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- 04-04-2011 19:48
i got the same answer as you
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SHUAIB_AKA_SHU
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- 04-04-2011 19:48
(Original post by soutioirsim)
Did you remember to divide by?
(If you use a u-sub of "u = cos2x" you end up having to divide by, you could of missed it out...)
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SHUAIB_AKA_SHU
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- 04-04-2011 19:50
(Original post by cccdude)
i got the same answer as you -
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- 04-04-2011 19:51
nah its not by substituion.. its through standard patters
let Y= (cos2x+3)^1/2
Differentiate to give -sin2x/(cos2x+3)^-1/2
Using the identity sin2x-2sinxcosx you will have to divide the function by 2 to get back to the original equation -
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- 04-04-2011 19:51
sin2x=2sinxcosx
Made an error in previous post -
SHUAIB_AKA_SHU
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- 04-04-2011 19:52
(Original post by cccdude)
nah its not by substituion.. its through standard patters
let Y= (cos2x+3)^1/2
Differentiate to give -sin2x/(cos2x+3)^-1/2
Using the identity sin2x-2sinxcosx you will have to divide the function by 2 to get back to the original equation -
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- 04-04-2011 19:55
nah the whole thing with this integration is kinda guess work.. you have the guess the differential as i did which was cos2x+3 to the power of 1/2.
you then differentiate and the end answer should be something like what you the question was asking but then has to be altered.. in our case we had to divide by a negative half
understand? -
soutioirsim
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- 04-04-2011 19:56
(Original post by SHUAIB_AKA_SHU)
im nt using tht method of u du etc. im jus integrating a section making the trig function then 4rm there.....group and intregrate. i dnt get wt i did wrong in my method lol :/
I got the right answer with a u sub though. -
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- 04-04-2011 19:58
its quite complicated but is needed on the edexcel C4 syllabus
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SHUAIB_AKA_SHU
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- 04-04-2011 20:01
(Original post by soutioirsim)
I don't get your method sorry
I got the right answer with a u sub though. -
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- 04-04-2011 20:03
RIP extricated
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- 05-04-2011 17:40
Can anybody show me how to integrate (sinx)*(x^2) dx ?
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Mr M
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- 05-04-2011 17:41
(Original post by diane_h)
Can anybody show me how to integrate (sinx)*(x^2) dx ? -
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- 05-04-2011 17:56
u=X^2
Du/Dx=2X
DX=DU/2X
X=SQRT(U) -
Mr M
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- 05-04-2011 18:00
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- 05-04-2011 18:06
(Original post by Mr M)
Er ... do you know what integration by parts is diane?
Int(U) dv = uv- INT(V) du?
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