Question

Help Pls!

∫(sinx)^4 dx

Answer 1

you can either use the reduction formulas, or repeatedly apply the identity\[\sin^2x=\frac12[1-\cos(2x)]\]

Answer 2

followed by\[\cos^2u=\frac12[1+\cos(2u)]\]

Answer 3

well I broke it down like this,

∫(sinx)^4=∫((sinx)^2)^2=∫((1+cos2x)/2)^2

Answer 4

then I got stuck...

Answer 5

expand that and what do you get?

Answer 6

wait wait, first of all that should be 1-cos(2x)

Answer 7

i belive it is 1/4∫1-2cos2x+(cosx)^2

Answer 8

the last term is wrong, why did you change the argument from cos(2x) to cosx?

Answer 9

right right it should be (cos2x)^2 correct?

Answer 10

right

Answer 11

so that term needs to be dealt with to be able to integrate it, so what formula can we apply here?

Answer 12

im drawing a blank here...

Answer 13

this is where i got stuck

Answer 14

we only have two to work with\[\sin^2u=\frac12[1-\cos(2u)]\]\[\cos^2u=\frac12[1+\cos(2u)]\]

Answer 15

"but wait" you say, "I have cos(2x), not cos(x), how can I apply the second formula?"
answer: let u=2x

Answer 16

ahhh....

Answer 17

so 1/8∫1-2cosu+(cosu)^2 ?

Answer 18

yeah, now how would you change (cosu)^2 ?

Answer 19

since you have cosu already would you change it to 1-(sinu)^2?

Answer 20

no, because how the heck do you integrate sin^2 ?
we want to get everything to the power of 1, so apply one of the double angle formulas I have written above

Answer 21

so (1+cos2u)/2

Answer 22

correct, now you can re-substitute 2x for u

Answer 23

i think i may have messed this up but is it 1+cos4x/2 ?

Answer 24

you only messed up in the lack of parentheses
(1+cos4x)/2 ?

Answer 25

right...

Answer 26

so now the whole thing is....?

Answer 27

1/16∫1-2cos2x+1+cos4x

Answer 28

you didn't simplify right
1/4∫1-2cos2x+(cosx)^2dx= 1/4∫1-2cos2x+(1+cos4x)/2dx=1/8∫2-4cos2x+1+cos4xdx=...

Answer 29

ok so the 1/2 is the same as multiplying the other terms by 2 then...

Answer 30

yes, if you factor 1/2 out of the integrand, then you factor 1/2 out of each term, which is the same as dividing each term by 1/2, and x/(1/2)=2x

Answer 31

ok, so now you can just push the integral inside and solve for each one then correct?

Answer 32

keeping the 1/8 with the integral for each one...

Answer 33

yep, now integrate term-by-term
I'd keep the 1/8 outside until the very end

Answer 34

ok i think i got it thanks...

Answer 35

welcome !

Answer 36

just curious are you an undergrad or graduate student?

Answer 37

I am neither. Completely self-taught, basically. Never been to college, so feel free to doubt me on anything :D

Answer 38

oh haha. well good job teaching yourself!

Answer 39

Thanks, I just happen to like this stuff. I hope to go to school eventually...