Question

find the integral

(cos(x))^9 sin(x)dx

Answer 1

ok so...

Answer 2

and what exactly is the problem here?
I've seen you do much harder than this

Answer 3

\[\sin(x)(\cos(x)^2)^4 \cos(x)\]

Answer 4

why?

Answer 5

why would you want to do that?

Answer 6

Just going by the section stuff :P.

Answer 7

Maybe it's easier than I'm thinking?

Answer 8

also\[\cos(x)^2\]is very confusing notation, does it mean\[\cos^2x\]or\[\cos(x^2)\]

Answer 9

and yes, it's way easier than you are thinking

Answer 10

standard table of integrals for
\[\int\sin ^nx\cos^mx\text dx\]

Answer 11

I guess (cos(x))^9 is the same as cos^9(x)?

Answer 12

right

Answer 13

what rhauskus said :P

Answer 14

but that is for the general case
your problem should be solvable by eye

Answer 15

I could say u = cos(x)
du = -sin(x)dx
-du = sin(x)dx

\[-\int\limits u^9 du\]

Answer 16

exactly :)

Answer 17

\[\frac{-u^{10}}{10} +c\]

Answer 18

\[\frac{-cos^{10}}{10} +c\]

Answer 19

-cos(x)

Answer 20

to make the cos part look better put a forward slash first
\cos{whatever}\[\cos{whatever}\]

Answer 21

\[\cos(x)\]

Answer 22

NEXT QUESTION!!!