1. integrals help

Compute an antiderivative for each of the following functions:

$\displaystyle a) f(x) = \frac{2x^3+1}{x^2}$
$\displaystyle b) g(x) = \sqrt{x}+x^\frac{7}{3}$
$\displaystyle c) h(x) = csc(2x)cot(2x)$

I was never taught integrals properly, only the basics where u change $\displaystyle x^2$ to $\displaystyle x^3/3$.

can anyone help? thanks!

2. Originally Posted by break
Compute an antiderivative for each of the following functions:

$\displaystyle a) f(x) = \frac{2x^3+1}{x^2}$
$\displaystyle b) g(x) = \sqrt{x}+x^\frac{7}{3}$
$\displaystyle c) h(x) = csc(2x)cot(2x)$

I was never taught integrals properly, only the basics where u change $\displaystyle x^2$ to $\displaystyle x^3/3$.

can anyone help? thanks!

For a polynomial $\displaystyle ax^n$ the integral is given by $\displaystyle a\int x^ndx=\frac{a}{n+1}x^{n+1}+C$. Convince yourself this is true by differentiating.

The first one can be written as a sum:

$\displaystyle \int\frac{2x^3+1}{x^2}dx=\int2xdx+\int\frac{1}{x^2 }dx$

Use the formula. Remember that $\displaystyle \frac{1}{x^2}=x^{-2}$.

Problem (b) just requires the formula. Remember that $\displaystyle \sqrt{x}=x^{\frac{1}{2}}$.

For (c) rememeber that $\displaystyle \frac{d}{du}csc(u)=-csc(u)cot(u)$

So if you let $\displaystyle u=2x$ so that $\displaystyle du=2dx$ then $\displaystyle csc(2x)cot(2x)dx=\frac{1}{2}csc(u)cot(u)du$. Does this make sense? Do you see what the integral is?

3. I had to edit my original post for latex error. Sorry about that.

The first one can be written as a sum:

$\displaystyle \int\frac{2x^3+1}{x^2}dx=\int2xdx+\int\frac{1}{x^2 }dx$

Use the formula. Remember that $\displaystyle \frac{1}{x^2}=x^{-2}$.
ok so this turns to

$\displaystyle \int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

$\displaystyle \int x^2dx + \int\frac{-1}{x}dx$

i think i did something wrong

5. Originally Posted by break
Compute an antiderivative for each of the following functions:

$\displaystyle a) f(x) = \frac{2x^3+1}{x^2}$
$\displaystyle b) g(x) = \sqrt{x}+x^\frac{7}{3}$
$\displaystyle c) h(x) = csc(2x)cot(2x)$

I was never taught integrals properly, only the basics where u change $\displaystyle x^2$ to $\displaystyle x^3/3$.

can anyone help? thanks!
Alternatively, for c), remember

$\displaystyle \csc{(2x)}\cot{(2x)} = \frac{1}{\sin{(2x)}}\cdot \frac{\cos{(2x)}}{\sin{(2x)}}$

$\displaystyle = \frac{\cos{(2x)}}{\sin^2{(2x)}}$.

Now use a $\displaystyle u$ substitution.

Let $\displaystyle u = \sin{(2x)}$ so that $\displaystyle \frac{du}{dx} = 2\cos{(2x)}$.

Then you have

$\displaystyle \int{\csc{(2x)}\cot{(2x)}\,dx} = \int{\frac{\cos{(2x)}}{\sin^2{(2x)}}\,dx}$

$\displaystyle = \frac{1}{2}\int{\frac{1}{\sin^2{(2x)}}[2\cos{(2x)}]\,dx}$

$\displaystyle = \frac{1}{2}\int{\frac{1}{u^2}\,\frac{du}{dx}\,dx}$

$\displaystyle = \frac{1}{2}\int{u^{-2}\,du}$

$\displaystyle = -\frac{1}{2}u^{-1} + C$

$\displaystyle = -\frac{1}{2}\cdot\frac{1}{\sin{(2x)}} + C$

$\displaystyle = -\frac{1}{2}\csc{(2x)} + C$.

6. Originally Posted by Prove It
Then you have

$\displaystyle \int{\csc{(2x)}\cot{(2x)}\,dx} = \int{\frac{\cos{(2x)}}{\sin^2{(2x)}}\,dx}$

$\displaystyle = -\frac{1}{2}\int{\frac{1}{\sin^2{(2x)}}[-2\cos{(2x)}]\,dx}$
i got lost here. where did u get the -1/2 from? is there a rule for this?

7. Originally Posted by break
ok so this turns to

$\displaystyle \int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

$\displaystyle \int x^2dx + \int\frac{-1}{x}dx$

i think i did something wrong
You're way off. The integral $\displaystyle \int 2xdx=\frac{2}{3}x^3+C$ is given by the formula $\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$. Do you see how I applied the formula? It works the same for all polynomials. See if you can do the same for the second term.

Here is a tutorial that might help you with some of these. It seems like you don't have the basic concept of what an integral or anti-derivative is.

Pauls Online Notes : Calculus I - Indefinite Integrals

8. Originally Posted by break
i got lost here. where did u get the -1/2 from? is there a rule for this?
1. Notice my edit.

2. We need to change the integrand so that it has $\displaystyle \frac{du}{dx}$ in it.

So we change $\displaystyle \cos{(2x)}$ into $\displaystyle \frac{1}{2}\cdot 2\cos{(2x)}$.

Then I took the $\displaystyle \frac{1}{2}$ outside the integral.

You're way off. The integral $\displaystyle \int 2xdx=\frac{2}{3}x^3+C$ is given by the formula $\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$. Do you see how I applied the formula? It works the same for all polynomials. See if you can do the same for the second term.

Here is a tutorial that might help you with some of these. It seems like you don't have the basic concept of what an integral or anti-derivative is.

Pauls Online Notes : Calculus I - Indefinite Integrals
why does x skip over $\displaystyle x^2$ and go to $\displaystyle x^3$? I see how the formula says n+1 but, 1+1 = 3? i don't get it.

10. Originally Posted by break
why does x skip over $\displaystyle x^2$ and go to $\displaystyle x^3$? I see how the formula says n+1 but, 1+1 = 3? i don't get it.
You're right

It should be $\displaystyle x^2+C$.

$\displaystyle \int 2xdx =\frac{2}{1+1}x^{1+1}+C=x^2+C$

Sorry about that. That was just a stupid mistake on my part.

You're right

It should be $\displaystyle x^2+C$.

$\displaystyle \int 2xdx =\frac{2}{1+1}x^{1+1}+C=x^2+C$

Sorry about that. That was just a stupid mistake on my part.
haha my brain was seriously about to explode right now. phew.

Originally Posted by break
$\displaystyle \int\frac{2x^2}{2} + \int\frac{x^-1}{-1}$

$\displaystyle \int x^2dx + \int\frac{-1}{x}dx$

i think i did something wrong
so i did get the first part right... right? haha. do i still have to plug in a + C early on?