Question

Integrate :|

Answer 1

\[\Huge \int\limits_{}^{} \frac{\cos^5x}{{sinx}}\]

Answer 2

Attempt:
Let sinx=t
dx=dt/cosx
\[\Huge \int\limits_{}^{} \frac{\cos^4x}{t}dt\]
wish we could change cos^4x in terms of t..dunno how

Answer 3

That one you'll have to break up for sure. You have to do something like:
\[\int\limits_{}^{}\frac{ (1-\sin ^{2}x)(1-\sin ^{2}x)(cosx) }{ (sinx) }\]

Answer 4

what now??

Answer 5

Multiply it out and turn it into separate fractions.

Answer 6

.......:/

Answer 7

what?

Answer 8

cant we change cos^4x in terms of t?

Answer 9

Okay so if we have t, then why is everything in x? Does the problem actually have cosx sinx but then dt?

Answer 10

i have made an assumption

Answer 11

can we write cos^4x as (1-t^2)^2?

Answer 12

Well I was just making the assumption that we had:
\[\int\limits_{}^{}\frac{ \cos ^{5}x }{ sinx }dx\]

Answer 13

yes we have that

Answer 14

Well then we would be doing what I am doing.

Answer 15

I guess you use t instead of u to substitute, though O.o

Answer 16

yeah

Answer 17

Well we would be multiplying it out like I was saying. There is no substitution yet, just manipulation.

Answer 18

\[\Huge \int\limits_{}^{} \frac{(1-t^2)^2}{t} dt\]
using my method^

Answer 19

Okay, your method. Where'd the extra cosx go? You had 5 of them originally in the numerator.

Answer 20

dx=dt/cosx

Answer 21

Wouldn't you still have to multiply that out anyway?

Answer 22

we can take t common from numerator

Answer 23

Well if you know what you're doing from here then yeah, just continue. Anything I'm going to do is going to conflict with your thinking. And it seems like you know what you're doing O.o

Answer 24

hhahha it becomes easy,expand and integrate individually

Answer 25

i got the answer :D

Answer 26

Same thing, you just wanted to do a substitution along with it O.o Meh, nvm, I'm just getting in your way -_-