Question

intergration problem,
how to do
intergration of [3sin^2(x/2)]^2

Answer 1

that is a mess aint it

Answer 2

9 sin^4(x/2) ; and pull out the 9 for later

\(\int\) sin^4(x/2) dx

Answer 3

eh got a typo
should be ∫ (3sin(x/2))^2 dx

Answer 4

sorry dude

Answer 5

sin^2 sin^2 ; where sin^2 = 1 -cos^2

sin^2(1-cos^2)

\(\int\) sin^2 - sin^2cos^2

Answer 6

........ ok

Answer 7

(3 sin(x/2))^2 = 9 sin^2(x/2) ; pull out the 9

9 \(\int\) sin^2(x/2) dx

9 \(\int\) \(\frac{2}{2}\)[sin(x/2)]^2 dx ; pull out the 2 on top

9.2 \(\int\frac{1}{2}\) [sin(x/2)]^2 dx

Answer 8

im gonna have ta write it on paper to keep track of it; typing and mathing dont go together :)

Answer 9

lol this is hard

Answer 10

Ok, here is how to do it.
First, do a substitution u=x/2 - just to make it a bit easier to write
Then, realize that sin^2=(1-cos(2x))/2
I think you can get it from there...

Answer 11

yeah, i was just recalling that ..... lol thnx

Answer 12

i look at the answer and its says
cos 2x = 1+2sin^2x gives sin^2x = (1-cos2x)/2
cos x = 1+ 2sin^2(x/2) gives sin^2(x/2) = (1-cos x)/2


MESSY

Answer 13

\[sin^2(x/2)=\frac{1}{2}-\frac{cos(x)}{2}\]

Answer 14

ummm i dont get the subsitution

Answer 15

Oh, yeah, I suppose you don't need it.

Answer 16

its really about the trib identity for cos(2x) = 1 - 2sin^2(x)

Answer 17

err... trig even lol

Answer 18

\[9\int\left(\frac{1}{2}-\frac{1}{2}cos(x)\right)\ dx\]

Answer 19

ummm thx
but can u explain in the u subsitution way?
coz i always get stuck in this method

Answer 20

there is no usub method for this .... its jusst noticinng that sin^2 is an identity and this is how you integrate it

Answer 21

Well...I suppose you could attempt to come up with a creative u, likely involving inverse trig functions, which will be integrable by basic methods, but would probably be a pain to write...

Answer 22

lol ok thx both of u ^^

Answer 23

youre welcome :)