Question

help?

Answer 1

let me write!

Answer 2

\[\int\limits sinx cosx dx\]

Answer 3

i give u choices wait..

Answer 4

\[ 1.\frac{ \sin^2x }{ 2 } 2. \frac{ \cos^2x }{ 2}\]

Answer 5

cos^2x/2

Answer 6

actually its sin^2x/2... sorry

Answer 7

so u bring the power down, decrease the power by 1 and multiply by the derivative of sinx... do u understand that?

Answer 8

in mathematical form plz..

Answer 9

?

Answer 10

i dont know how to explain it in a better way... think of it this way: (sinx)^2 so u have to multipy it with the derivative of wats inside the brackets after u decrease the power... do u get it now?

Answer 11

it's really slow sorry for late reply

Answer 12

err... m sorry, m quite bad with explaining i hope some1 else can give u hand with it

Answer 13

it's okay :)

Answer 14

sin x cos x = 1/2 * sin(2x) so just integrate 1/2 sin (2x)

Answer 15

you should get -1/2 cos^2 (x)

Answer 16

alternatively you could let u = sin x, and du = cos x. or let u = cos x and let du = -sin x.. so many ways

which way do you prefer

Answer 17

answer should be sin^2x/2...!

Answer 18

?

Answer 19

\[\int\limits_{}^{} \sin x * \cos x dx = \frac{ \sin^{2} x }{ 2 }\]

?

Answer 20

yeah

Answer 21

ah ok
well i guess they only differ by a constant

\[\frac{ \sin^{2}x }{ 2 } = \frac{ -\cos^{2} x }{ 2 } + \frac{ 1 }{ 2 }\]

there are many answers to this question

Answer 22

thanks:)

Answer 23

u substitution
u=sinx
du=cosx dx
integral = udu
integrate that and you get (u^2)/2
Insert sinx back in for u and you get
[sin^2 (x)] / 2