Question

Can you help me solve this integral: sin^2t*cos^2t dt

Answer 1

you can rewrite it if you want using trig identities, might make it simpler

Answer 2

\[s^2(t)c^2(t)=(s(t)c(t))^2=s^2(\frac12t)\]
if i recall that correctly

Answer 3

err, sin(t)cos(t) = 1/2 sin(2t)

soo close

Answer 4

are you sure? i dont understand about last par

Answer 5

if you dont understand that identity then maybe:

cos^2=1-sin^2

therefore

sin^2 cos^2 = sin^2(1-sin^2) = sin^2 - sin^4

Answer 6

the integrals can be worked out by hand using the by parts method, or by table ....

Answer 7

so we have integral sin^2 2t dt, right?
and 1/4 before integral

Answer 8

yes, if thats the route you want to take :)

Answer 9

that's right, but before or after is same :)

Answer 10

\[\frac14\int sin^2(u)~du\]if you want to clean it up some

Answer 11

u = 2t
du = 2 dt

dt = du/2 sooo

\[\frac18\int sin^2(u)du\]

Answer 12

gotta run, good luck

Answer 13

but what to do with sin^2?
i dont know..

Answer 14

sin^2 is a basic integral by parts, which is also tabled up

Answer 15

oo, ok, thank you :)

Answer 16

or u can use the identity :
sin^2 (2t) = (1 - cos(4t))/2

Answer 17

thank you too :)