Question

integral problem please help

Answer 1

\[\int\limits_{?}^{?} \cos ^{2}x sinx dx\]

Answer 2

i am stuck can you help me

Answer 3

you have to use the substitution rule.

Answer 4

so what would you let u=?

Answer 5

u=cos

Answer 6

what would du =?

Answer 7

du=-sin

Answer 8

Okay, so u=cosx, and du=-sinxdx, don't forget the x's and dx lol :P but we have a problem... in the question given, we have sinxdx, not -sindx, so what would you do to solve that?

Answer 9

-sinxdx*

Answer 10

i dont know

Answer 11

\[-\int\limits_{?}^{?} u du\] this?

Answer 12

you do know... but it you really think you don't know then i'll explain it to you so you can understand a bit more.. so we have u=cosx du=-sinxdx, in the problem we have \[\int\limits_{?}^{?}\cos ^{2}xsinxdx\], so we want there to be a negative -sinx, so what we do is multiple the integral by a negative, like you did. But we can't change the question, so we met two negatives...\[-\int\limits_{?}^{?} \cos ^{2}x*-sinxdx\]

Answer 13

do you see where i put the negatives?

Answer 14

yes i see

Answer 15

i put two negatives to make sure we didn't change the question.

Answer 16

so we dont use the sum difference of sine and cosine in this problem?

Answer 17

so when you subbed in u and du you kinda messed up.. u=cosx right? but we have cos^2x in the equation, so what would u be in the problem

Answer 18

im not sure about that, I'm pretty sure subing would be easier.

Answer 19

it would be \[-\int\limits_{?}^{?}u ^{2}du\]

Answer 20

and then you just solve the integral

Answer 21

okay thank you

Answer 22

tell me what you get, so we can compare :P