Question

can someone show the solution to the integral of sin^3 4x cos^5 4x dx

Answer 1

cos^5 can be converted into sin^3*cos^?

Answer 2

strip out a sin(4x) and you get\[\cos^5(4x)\sin^2(4x)\sin(4x)\]now use\[\sin^2u=1-\cos^2u\]and then a simple -sub

Answer 3

u-sub

Answer 4

amistre doesn't like my answer :(

Answer 5

does my u=cos 4x?

Answer 6

\[ sin^3(4x) cos^5(4x)\]
\[ sin^3(4x) (cos^2(4x))^2cos(4x)\]
\[ sin^3(4x)(1-sin^2(4x))^2cos^3(4x)\]
\[ sin^3(4x)(1-2sin^2(4x)+sin^4(4x))cos^3(4x)\]
\[ (sin^3(4x)-2sin^5(4x)+sin^7(4x))cos^3(4x)\]

and that is just alot of int f(x)f'(x) stuff

Answer 7

yes

Answer 8

if you do it my way
amistre's is clearly different

Answer 9

i undid the cos part, i think you undid the sin?

Answer 10

yes the rule said if both odd its either way

Answer 11

and i got a ^3 clingon lol

Answer 12

yeah\[\cos^5(4x)\sin^2(4x)\sin(4x)=\cos^5(4x)(1-\cos^2)\sin(4x)\]\[=\cos^5(4x)\sin(4x)-\cos^7(4x)\sin(4x)\]at least that's how mine unfolds...

Answer 13

i spose the smaller odd is a little easier to play with

Answer 14

\[u=\cos(4x)\implies du=-4\sin(4x)dx\]

Answer 15

just a little maybe....

Answer 16

always good to see more than one way though

Answer 17

my final answer is -cos^6 4x/24 + cos^8 4x/56 + c...please correct my mistakes..

Answer 18

not sure about the 56

Answer 19

yeah its 32, right?

Answer 20

after the sub we have a 1/4 in front of it, right?
than we divide by the exponent, which is...?

Answer 21

\[u=\cos(4x)\implies du=-4\sin(4x)dx\implies-\frac14du=\sin(4x)dx\]

Answer 22

oh yeah, that's 32, sorry I missed that comment

Answer 23

thanks turning test..where you from?

Answer 24

I am from the United States but live in Mexico
you?

Answer 25

Philippines..i'm studying bse math..

Answer 26

Cool
well, you're welcome for the help and I hope to see you around!