Question

find the integral.....

Answer 1

\[\int\limits \frac{(\cos(t))^5}{\sqrt{\sin(t)}} dt\]

Answer 2

\[\int\limits (\cos(t))^5\sin(t)^{1/2} dt\]

Answer 3

\[\int\limits (\cos^2(t)^2)\sin(t)^{1/2} \cos(t) dt\]

Answer 4

\[\int\limits (1-\sin^2(t))^2 \sin(t)^{1/2} \cos(t) dt\]
\[\int\limits (\sin(t)^{1/2}-\sin(t))^2 \cos(t) dt\]

\[\int\limits (\cos(t) \sin(t)^{1/2}-\sin(t)\cos(t))^2 dt\]

Answer 5

I don't think I'm doing this correctly...

Answer 6

\[\cos^2(x)^2\]means nothing, so be careful in your notation

Answer 7

one thing we can always try with these kind of integral is a u-sub that we can write in terms of the other trig function as per sin^2+cos^2=1

Answer 8

for instance\[u=\cos x\implies du=-\sin xdx=-\sqrt{1-\cos^2x}dx=-\sqrt{1-u^2}dx\]maybe that one won't pan out, but you see where I'm going hopefully

Answer 9

hmmm.

Answer 10

the book does a poor job explaining what to do with higher level exponents...

Answer 11

but i guess that makes sense so it will cancel out into 1-cos^2(x)?

Answer 12

where does that extra '-' come from?

Answer 13

oh nvm...

Answer 14

maybe (I'm just trying things) \[u=\sqrt{\sin t}\implies du=\frac{\cos tdt}{\sqrt{\sin x}}\]

Answer 15

u'/u = log|u| + c .......... ok next question!

Answer 16

now where to retrieve the other cos^4 is the problem

Answer 17

hmm so we are basically going to start off with u sub from the start huh..

Answer 18

This is stooooooooooopid :(.

Answer 19

lemme see how the book goes about it...

Answer 20

\[u=\sqrt{\sin t}\implies u^4=1-\cos^2t\implies\cos^4t=(1-u^4)^2\]ugly ugly...

Answer 21

next! :P

Answer 22

In the book for cos^4(x) they show using a half angle formula and then foiling it out....

Answer 23

so could we do that to make it extra ugly?

Answer 24

so that would be\[\int(1-u^4)^2du\]hey it's not that bad!

Answer 25

expand and integrate...

Answer 26

oh I think I forgot a 2...

Answer 27

isn't 1-c0s^2 = sin^2?

Answer 28

oh woops that's an arrow :P

Answer 29

\[\int\limits (1-u^4)(1-u^4)du\]

Answer 30

\[\int\limits (1-2u^4 + u^8)du??\]

Answer 31

\[u=\sqrt{\sin t}\implies u^4=1-\cos^2t\implies\cos^4t=(1-u^4)^2\]\[u=\sqrt{\sin t}\implies du=\frac{\cos tdt}{2\sqrt{\sin t}}\implies2du=\frac{\cos tdt}{\sqrt{\sin t}}\]so\[\int{\cos^5tdt\over\sqrt{\sin t}}=2\int(1-u^4)^2du\]so yes with a 2 in front

Answer 32

\[2(u-\frac{2u^5}{5} + \frac{u^9}{9}) + c??\]

Answer 33

This is stupid ugly.

Answer 34

but seems correct to me
let me ask the wolf for a second opinion (remember to do the sub)

Answer 35

\[2u-\frac{4u^5}{5} + \frac{2u^9}{9} + c??\]

Answer 36

\[2\sqrt{\sin}-\frac{4\sqrt{\sin}^5}{5} + \frac{2\sqrt{\sin}^9}{9} + c??\]

Answer 37

better to write it with fractional exponents

Answer 38

\[2\sqrt{\sin}-\frac{4\sin^3}{5} + \frac{2\sin^7}{9} + c??\]

Answer 39

Idk if I can do that :p

Answer 40

this is a really hard q... :(

Answer 41

\[2\sin^{1/2}t-{4\sin^{5/2}t\over5}+{2\sin^{9/2}t\over9}+C\]in fractional exponents it would be this

Answer 42

remember\[\sqrt x=x^{1/2}\]

Answer 43

stupid wolfram has "simplified" our answer beyond my ability to recognize if it is the same or not
http://www.wolframalpha.com/input/?i=integral%20cos%5E5x%2F(sqrt(sin%20x))dx&t=crmtb01

Answer 44

click "show steps" though and everything is the same up until the part that says "is equal to..." lol!

Answer 45

so I'm gonna call my answer correct

Answer 46

:)

Answer 47

LOL..

Answer 48

I gotta go, good luck on the rest dude :)

Answer 49

Thanks :). Lets hope it's correct or else :'(.

Answer 50

like I say, hit "show steps" and you see it has the same answer up until the very last step, but it figured a different way to wrote it is all

Answer 51

so I'm quite confident you're okay
see ya!

Answer 52

WOOT! :)

Answer 53

omfg soo stupid I hate these q's :'(

Answer 54

cya thaks for everything!!!

Answer 55

no prob

Answer 56

Still trying to see this, it's kinda complicated but I guess not too bad :P.