Question

∫sin^3(x)

Answer 1

\[\int\limits_{}^{}(\sin x)^{3}dx\]\[\int\limits_{}^{}(\sin x)^{2}(\sin x) dx\]\[\int\limits_{}^{}(1-(\cos x)^{2})(\sin x )dx\]\[\int\limits_{}^{}sinx dx -\int\limits_{}^{}(\cos x)^{2}(\sin x) dx\]

Answer 2

i get up to\[\int\limits (1-(\cos^2x))\sin^2x dx\] but then i get suck how did u find the rest?

Answer 3

the 1-(cos x)^2 can only come from (sin x)^2... which only leaves sin x for the second factor.

Answer 4

yes i get that, from trig identities. did u use u substitution? if so what did u use for u?

Answer 5

just be careful... yours doesn't match mine...

Answer 6

\[\int\limits_{}^{}(\sin x)^{3}=\int\limits_{}^{}(\sin x)^{2}(\sin x) dx\]

Answer 7

ya thats rewriting it. how does the one plus cosx squared go away?

Answer 8

one minus, correction

Answer 9

\[(\sin x)^{2}+ (\cos x)^{2}=1\]
say u= cos x... then du = - sin x dx

Answer 10

the u integral becomes
\[-\cos x +\int\limits_{}^{}u ^{2}du\]

Answer 11

look at my integral steps above... when I get to the third integral... I distribute the sin x through the one minus cosine squared on the left.

Answer 12

i dont see why u distribute. dont u want to get rid of sinx with du? then making it a u form and integrating?

Answer 13

by distributing, I make the first integral simple... \[\int\limits_{}^{}\sin x dx = -\cos x +c\]then the second integral requires a u-sub

Answer 14

The ultimate point of maniputlating integrals is to make a bunch of integrals that are each memorized integrals that you know the answer to.

Answer 15

so the answer is -(cosx) 1/3 (cosx)^3 ???

Answer 16

very close... there is a plus between the first cox x and the (1/3). and there is a +C on the end.

Answer 17

-cos x +(1/3)(cos x)^3 + C

Answer 18

its a plus because of the negative in front of the integral and du being negative sinx?

Answer 19

yes! perfect!

Answer 20

let me ask u this, how do u know when to stop manipulating and to integrate?

Answer 21

in general

Answer 22

You stop when you see something that you can do... i.e. simple integrals that are solved in one step, or integrals that can be solved with a simple two step solution such as u-sub or by parts... partial fraction decomposition and trigonometric substitution are different animals.

Answer 23

haha ok, thanks alot!

Answer 24

you're welcome... loads of fun!