Question

integrate sin^3x cos^5x dx

Answer 1

\[\int \sin^3(x)\cos^5(x)dx=-\frac{\cos^6(x)\sin^2(x)}{8}-\frac{\cos^6(x)}{24}\]

Answer 2

i cant believe you are here satellite it is early

Answer 3

yes looks good :)

Answer 4

damn wrong file

Answer 5

that is the right one sorry

Answer 6

which one its right?

Answer 7

i know my is right i haven't checked satelliltes

Answer 8

so qs. when i started doing it i broke up the cos^5x because i have understood that odd exponents u break and use substitution.... what is the exception in this excersice

Answer 9

? do you not understand what i did

Answer 10

no

Answer 11

i broke up sin^2x

Answer 12

u broke uo sin^3x

Answer 13

so you don't know sin^3x=sin^2x*sinx

Answer 14

and rewrite it as cosx

Answer 15

right

Answer 16

yea thats perfect

Answer 17

i understood that part

Answer 18

i am talking about cos^5x

Answer 19

i thought u do the samething

Answer 20

you can try it that way if you want

Answer 21

oh ok.. should give me the same thing?

Answer 22

yes if it works

Answer 23

yes that will work to i will show you both ways k?

Answer 24

yea,,, i finished mine.. :) it gave me the same

Answer 25

i started doing it but i got confuse... maybe i was wrong...

Answer 26

why did you get confuse

Answer 27

because it was long?

Answer 28

by the way satellite is wrong i check his and did not get sin^3xcos^5x back when i took derivative of his anti derivative

Answer 29

yea mine was longer...lol complicated....

Answer 30

oh wait he might be right too i took my algebra alittle bit further if i didn't make a mistake his answer is good

Answer 31

so how can u get two different answers?

Answer 32

it is the same answer

Answer 33

just like if we simplifed 6/2 to 3

Answer 34

do you remember trig identities

Answer 35

yes

Answer 36

there is alot of ways to right one thing

Answer 37

oh ok...

Answer 38

well because i did not see the step that he follow i prefer stick with this ones,,, later when i fell more confortable with this way i can try other... maybe is easier

Answer 39

he might had used a formula

Answer 40

ohhhhhhhhh
jijij,,

Answer 41

i don't know of the formula or if there is one i could look up and see

Answer 42

noo,,, i like this way for now.. thanks u myininaya.. very helpful

Answer 43

i dont like formulas

Answer 44

too easy to use

Answer 45

i dont like neither

Answer 46

i am good with numbers,,,, (logic) no with memorization

Answer 47

by the way i see those rules you are mentioning now

Answer 48

since both were odd it didn't matter what substituition we chose (u=sinx or u=cosx)
if the odd was on the cosine though and not the sine, then use u=sinx
if the odd was on the sine and not the cosine, then use u=cosx
if both powers were even, then you will have to use some trig identities

Answer 49

yep yep there is a formula

Answer 50

oh jijijii

Answer 51

this formula would be ugly for really high powers

Answer 52

u found it?

Answer 53

yep

Answer 54

i was trying to find a link just in case you wanted to look at it because i didn't want to type it i am still looking online for it, but i have in my cal book in the back

Answer 55

\[\int\limits_{}^{}\sin^n(u)\cos^m(u)du=\frac{\sin^{n-1}(u)\cos^{m+1}(u)}{n+m}+\frac{n-1}{n+m}\int\limits_{}^{}\sin^{n-2}(u)\cos^{m}udu\]

Answer 56

In a general case to integrate stuff of the form sin^nx cos^mx ,
1- if both m and n are odd, substitute either sin x=t or cos =t and then integrate.
2- if one is odd and the other is even, sustitute the even power, for eg if m is odd put sin x=t
3- if both are even, dont substitute you can solve by trignometric identities.