Question

Why is this answer wrong:

Answer 1

Here are my steps and my answer:

Answer 2

\[\int\limits_{}^{}\sin ^{5}xdx\]

Answer 3

= \[\int\limits_{}^{}\sin x(1-\cos ^{4}x)\]

Answer 4

U sub:
u = cosx
du=-sinxdx -> -du=sinxdx

Answer 5

intsinx=-cosx
then -sinxcos^4x=1/5cos^5x

so all together you need 1/5cos^5x-cosx+C

Answer 6

\[-\int\limits_{}^{}1-u ^{4}\] = \[-u+u ^{5}/5\]

Answer 7

So my final answer would be:

Answer 8

yep

Answer 9

\[−cosx+\cos5x/5 + c\]

Answer 10

But the answer give on pauls notes is not the same?

Answer 11

what does it say on there?

Answer 12

So is my methodology incorrect or did it just yield a different form of the correct answer

Answer 13

http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx

Answer 14

example 1

Answer 15

he did a different trig. sub.

Answer 16

ok do you see where he does sin^5x=(1-u^2)^2 is wrong he is missing something cause that would yeild sin^4x

Answer 17

He has the other sinx include outside of that parenthesis

Answer 18

If you look before the u sub

Answer 19

sin^4x does not equal 1-cos^4x thats what you did wrong he is rightt

Answer 20

Why not? isnt that the fundamental trig.

Answer 21

ohhh yea

Answer 22

it only works for sin^2x+cos^2x

Answer 23

1-2cos^2x_cos^4x

Answer 24

Gotta remember that, thanks mate

Answer 25

so its easy once you do that but THE INTEGRAL YOU DID BEFORE THIS IS CORRECT KEEP WORKING HARD THESE ARE FAILY SIMPLE ONCE YOU GET THE HANG OF THEM