C4 integration....help!!!

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 :/

Reply 1

SHUAIB_AKA_SHU

OP

.......

Reply 2

soutioirsim

15

Did you remember to divide by

\frac{1}{2}

?

(If you use a u-sub of "u = cos2x" you end up having to divide by

\frac{1}{2}

, you could of missed it out...)

Reply 3

SHUAIB_AKA_SHU

OP

Original post by soutioirsim

Did you remember to divide by

\frac{1}{2}

?

yh 0.5.....

Reply 4

cccdude

i got the same answer as you

Reply 5

SHUAIB_AKA_SHU

OP

Original post by soutioirsim

Did you remember to divide by

\frac{1}{2}

?

(If you use a u-sub of "u = cos2x" you end up having to divide by

\frac{1}{2}

, you could of missed it out...)

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 :/

Reply 6

SHUAIB_AKA_SHU

OP

Original post by cccdude

i got the same answer as you

lol! wt method....reverse chain rule etc.?

Reply 7

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

Reply 8

cccdude

sin2x=2sinxcosx

Made an error in previous post

Reply 9

SHUAIB_AKA_SHU

OP

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

well i jus did it via integrating a section....making it the same trig then integratin 2geva.....:/

Reply 10

cccdude

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?

Reply 11

soutioirsim

15

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 :/