Question

Can someone help explain how to do this integral and get the answer?

Answer 1

\[\int\limits_{}^{}\cot^5xsin^2xdx\]

Answer 2

general idea is to change everything in terms of sin cos parts; then cancel what can be canceled

Answer 3

how would i do that?

Answer 4

waht another way to write cot?

Answer 5

yes

Answer 6

\[cot\ne yes\]

Answer 7

??

Answer 8

write cot in terms of its sin cos parts .....

Answer 9

Lol

Answer 10

you are mean

Answer 11

bye. no medals for you!

Answer 12

we dont do this for the medals, we do it to help people study the material.

can you rewrite cot in its sin cos parts?

Answer 13

no. i don't know what you mean!

Answer 14

there are two basic trig functions, sin and cos

these function can be used to define every other trig function, and they are taught to a student before they take calculus 1

Answer 15

i know that. i have no idea what to do with that

Answer 16

by redefining the cot in its sin cos parts, we can simplify the integrand to something more workable

Answer 17

\[cot^p~nsin^q=\frac{cos^p~sin^q}{sin^p}=cos^p~sin^{q-p}\]
this is hopefully a simpler integrand since cos and sin are derivatives of each other

Answer 18

got a spurious "n" in that ....

Answer 19

ok but what do i separate? can you like give me the first step

Answer 20

that was the first step
\[cot=\frac{cos}{sin}\]
using this in the integral will simplify it to:\[\int cos^5(x)~sin^{2-5}(x)dx\]
\[\int cos^5(x)~sin^{-3}(x)dx\]
still thinking about how to go from there tho

Answer 21

the wolf does not give a pretty result ......

Answer 22

looks like it would have to go thru some "by parts"

Answer 23

this is getting messy ....

hmmm, try a usub maybe?

u = sin^2x

du = 2 sinx cosx dx

du/2sinx cosx = dx

\[\frac12\int \frac{cos^5}{sin^5~sin~cos}\]
\[\frac12\int \frac{cos^4}{sin^6}du\]
\[\frac12\int cot^4u~csc^6u du\]
this at least gets us to derivative parts ....

Answer 24

and from there its still a bunch of "by parts" calculating ....