# calc 1 sin(x^3) integral

• Nov 22nd 2008, 02:35 PM
littlejodo
calc 1 sin(x^3) integral
I've searched online for an answer to this, but the answers I've found are full of (gamma?) notation that I don't understand and haven't yet been taught. I didn't think it should be that difficult, yet I can't figure out the answer.

Evaluate the integral of sin(x^3) dx

I know the antiderivative of sin is -cos, and the antiderivative of x^3 could be x^4 / 4 but I can't figure out how to make it work here.

I tried making u = x^3 so that the answer would be -cos u^2/2 but then is the answer -1/2cos(x^3)^2 ??

Surely I should have to integrate x^3, right?
• Nov 22nd 2008, 02:51 PM
galactus
Basic u subs will not work too well for this. It is not easily doable by

elementary means. That is why you found it in terms of gamma and what not.

If you are studying calculus, perhaps you mean

$\displaystyle \int sin^{3}(x)dx$ and not $\displaystyle \int sin(x^{3})dx$

Two entirely different animals.

That is probably the case.

Go to the Integrator and check it out at http://integrals.wolfram.com/index.jsp

Type in sin[x^3] and it will show you it ain't pretty. That is why I think it is the other.
• Nov 22nd 2008, 03:03 PM
galactus
If it is indeed $\displaystyle \int sin^{3}(x)dx$, then rewrite as:

$\displaystyle \int sin^{2}(x)sin(x)dx$

$\displaystyle \int (1-cos^{2}(x))sin(x)dx$

$\displaystyle \int sin(x)dx-\int cos^{2}(x)sin(x)dx$

Now, simple subs will work. If this is the case afterall.
• Nov 22nd 2008, 03:39 PM
littlejodo
unfortunately... very unfortunately, I entered it correctly.

I need the integral of sin(x^3)

any pointers?

I have to find this because ultimately I need the definite integral (1 on top and t on the bottom) and then I have to find the derivative of this.

so, F(t) = (1 on top, x on bottom of integral sign) of sin(x^3) dx

Find F'(x)

I tried jumping around the integrating just to find the derivative part and entered: sin(1) - sin(x^3) but it wasn't correct. I was hoping it was a shortcut.
• Nov 22nd 2008, 03:46 PM
galactus
For goodness sake, why didn't you say that in the first place?. That makes all the difference. It's the second fundamental rule of calc. There is no need to laboriously find the integral.

Is this what it is:

$\displaystyle \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=f(x)$

For instance, $\displaystyle \frac{d}{dx}\left[\int_{1}^{x}\frac{sin(t)}{t}dt\right]=\frac{sin(t)}{t}$

Remember that $\displaystyle \int_{a}^{x}f(t)dt=-\int_{x}^{a}f(t)dt$
• Nov 22nd 2008, 03:49 PM
littlejodo
But then why didn't sin(1) - sin(x^3) work? Oh... I was supposed to just enter - sin(x^3). I'm sorry.

I was on the right track to begin with. I hate online submission.
• Nov 23rd 2008, 12:01 AM
Moo
Quote:

Originally Posted by galactus
For goodness sake, why didn't you say that in the first place?. That makes all the difference. It's the second fundamental rule of calc. There is no need to laboriously find the integral.

Is this what it is:

$\displaystyle \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=f(x)$

For instance, $\displaystyle \frac{d}{dx}\left[\int_{1}^{x}\frac{sin(t)}{t}dt\right]=\frac{sin({\color{red}x})}{{\color{red}x}}$

Remember that $\displaystyle \int_{a}^{x}f(t)dt=-\int_{x}^{a}f(t)dt$

Just being picky (Tongueout)

To the OP : why isn't there sin(1) ?
Imagine that $\displaystyle F$ is an antiderivative of your function f.

We know that $\displaystyle \int_a^x f(t) ~ dt=F(x)-F(a)$

But a is a constant with respect to x. So F(a) is a constant too.
Hence when differentiating, you'll only have $\displaystyle F'(x)$, which is exactly $\displaystyle f(x)$