Question

I'm trying to integrate sin^3x, so I broke it up into sin^2x(sinx) Then I used the half-angle formula for sin^2x and got sin(x)/2-cos(2x)sinx.

I tried integrating that way and got (1/6)cos(3x) -cosx, but apparently the correct answer is (1/12)(cos3x -9cosx).

Can someone tell me why this is?

Answer 1

Try using the Pythagorean theorem: \[\int \sin^3 x dx = \int \sin^2 x \sin x dx = \int (1-\cos^2x)\sin x dx\] Perform u-substitution: \[\text{Let }u=\cos x \text{, thus } du=-\sin x dx \]\[-\int (1-u^2)du=-\left(u-\dfrac{u^3}{3}\right)+C = \dfrac{\cos^3 x}{3} - \cos x+C\] It's still the right answer but just a lot cleaner method of integration in my opinion. You don't have to deal with double or triple angles.

Answer 2

Okay, but I used Wolfram Alpha, and got this

http://www.wolframalpha.com/input/?i=sin^3x

If you scroll down to the indefinite integral part, you will see it is not the same.

Answer 3

just because something doesn't look the same doesn't mean they aren't the same

Answer 4

Then please tell me how cos(3x)/12 -(3/4)cosx = cos(3x)/6 - cosx.

Answer 5

prove it using trig identities

Answer 6

There is no using trig identities man, it's just 1/12 != 1/6 and 3/4 != 1.

Answer 7

Be careful! It is not that \[\dfrac{\cos(3x)}{12} -\dfrac 34 \cos x = \dfrac{\cos(3x)}{6} - \cos x\] but rather it's that\[\dfrac{\cos(3x)}{12} -\dfrac 34 \cos x + C_1 = \dfrac{\cos(3x)}{6} - \cos x + C_2\]where\[C_1,\text{ }C_2 \in \mathbb R\] (i.e. are independent constants) Have fun! :-)

Answer 8

\[(\frac{1}{3}\cos^3(x)-\cos(x)+C)'=\frac{1}{3} \cdot 3 \cos^2(x) (-\sin(x))-(-\sin(x))+0\]
\[-\cos^2(x)\sin(x)+\sin(x)= \]
\[\sin(x)(-\cos^2(x)+1)=\sin(x)(1-\cos^2(x))=\sin(x)(\sin^2(x))=\sin^3(x) \]

Answer 9

we get what we started with when we take derivative of our anti derivative so it is correct

Answer 10

Okay, but say it's a definite integral from 0 to pi. Plugging that in, wouldn't I have
-(1/12)+(3/4+(1/12)+(3/4) = 1.5
and
-(1/6)+1+(1/6)+1 = 2 ?

Is this allowed?

Answer 11

you will get output 4/3 for either one
using wikepedia's you get 4/3
using the one above you get 4/3

Answer 12

wikepedia lol!

Answer 13

whatever its called

Answer 14

just teasing. i wanted to see you show that this bunch of trig stuff = that bunch of trig stuff

Answer 15

and i did that above in that little file :)

Answer 16

No I"m not talking about your answer, I'm talking about the ones I posted.

Answer 17

\[\frac{1}{3}\cos^3(\pi)-\cos(\pi)-[\frac{1}{3}\cos^3(0)-\cos(0)]=\frac{1}{3}(-1)^3-(-1)-\frac{1}{3}(1)^3+1\]
\[=-\frac{1}{3}+1-\frac{1}{3}+1=2-\frac{2}{3}=\frac{6}{3}-\frac{2}{3}=\frac{4}{3}\]

Answer 18

I don't think my original method of splitting the sin^3x apart to sin^2x(sinx) works. Though I'm not sure why.

Answer 19

The method that got the (1/12) version, is using the reduction formula.

Answer 20

\[\int\limits_{}^{}\sin^3(x)dx=\int\limits_{}^{}\sin^2(x)\sin(x)dx=\int\limits_{}^{}\frac{1}{2}(1-\cos(2x))\sin(x) dx\]
\[\int\limits_{}^{}\frac{1}{2}(\sin(x)-\sin(x)\cos(2x)) dx=\int\limits_{}^{}\frac{1}{2}\sin(x) dx-\int\limits_{}^{} \frac{1}{2}\sin(x)\cos(2x) dx\]
\[\frac{1}{2}(-\cos(x))-\frac{1}{2}\int\limits_{}^{}\sin(x)(\cos^2(x)-\sin^2(x)) dx+C\]
\[=\frac{-\cos(x)}{2}-\frac{1}{2}\int\limits_{}^{}\sin(x)\cos^2(x)+\frac{1}{2}\int\limits_{}^{}\sin^3(x) dx+C\]
subtract \[\frac{1}{2}\int\limits_{}^{}\sin^3(x) dx\] on both sides
\[\int\limits_{}^{}\sin^3(x) dx-\frac{1}{2}\int\limits_{}^{}\sin^3(x) dx=\frac{-\cos(x)}{2}-\frac{1}{2}\int\limits_{}^{} \sin(x)\cos^2(x)+C\]
\[(1-\frac{1}{2})\int\limits_{}^{}\sin^3(x)dx=-\frac{\cos(x)}{2}-\frac{1}{2}\int\limits_{}^{}\sin(x)\cos^2(x)+C\]
we need to integrate
\[\int\limits_{}^{}\sin(x)\cos^2(x)dx\]
let \[u=\cos(x) => du=-\sin(x) dx=>-du=\sin(x) dx\]
\[\int\limits_{}^{}u^2(-du)=-\int\limits_{}^{}u^2du=-\frac{u^3}{3}+c=-\frac{\cos^3(x)}{3}+c\]
so back to
\[(1-\frac{1}{2})\int\limits_{}^{}\sin^3(x)dx=-\frac{\cos(x)}{2}-\frac{1}{2}\int\limits_{}^{}\sin(x)\cos^2(x)+C\]
\[\frac{1}{2}\int\limits_{}^{}\sin^3(x)dx=-\frac{\cos(x)}{2}+\frac{1}{2} \cdot \frac{\cos^3(x)}{3}+C\]
now multiply 2 on both sides
\[\int\limits_{}^{}\sin^3(x) dx=-\cos(x)+\frac{\cos^3(x)}{3}+C\]

Answer 21

this is the way you wanted to do it
but using
\[\sin^2(x)=1-\cos^2(x)\] is the easier way

Answer 22

Thanks, it makes sense now. Also, what do I do if I need to integrate higher powers of sine and cosine?

Answer 23

\[\int\limits_{}^{}\sin^m(x)\cos^n(x) dx\]

if n id odd, then rewrite cos^n(x)

if m is odd, then rewrite sin^m(x)

if both are even, use half-angle identities or double angle identities

Answer 24

In addition to above, if m or n equal 1, go directly to u substitution. (duh)