Why does int[2]dx = 2x, int[sin(x)]dx = -cos(x), etc?

[gopher]

so we finally started integration in class
I wonder why
\(\displaystyle \displaystyle\int 2\,dx = 2x\)
and
\(\displaystyle \displaystyle\int sin{x}\,dx = -cos{x}\)
why
\(\displaystyle \displaystyle\int 2*sin{x}\,dx = -2*cos{x}\)
and not
\(\displaystyle \displaystyle\int 2*sin{x}\,dx = 2x*-cos{x}\)?

 

[skeeter]

\(\displaystyle \L \int f(x) \cdot g(x) dx \neq \int f(x) dx \cdot \int g(x) dx\)

 

[stapel]

Did your class not cover anti-derivatives at all? (It would highly unusual to get to integrals without having first done anti-derivatives, and anti-derivaties would have answered your questions, is why I ask.)

Thank you.

Eliz.

 

[ChaoticLlama]

Because you are placing too much importance on the '2'.
Don't worry, you merely are having difficulty with notation.

\(\displaystyle \displaystyle\int 2\,dx
= 2\int\,dx
= 2x\)
The function you are integrating here is '1'

\(\displaystyle \displaystyle\int sin{x}\,dx = -cos{x}\)
The function you are integrating here is 'sin(x)'

\(\displaystyle \displaystyle\int 2*sin{x}\,dx
= 2\int sin{x}\,dx
= -2*cos{x}\)
the function you are integrating here is still 'sin(x)'

However skeeter's reply is the most concise.
If your logic were true, then doing integrals like \(\displaystyle \displaystyle\int e^x sin{x}\,dx\) would be doable in one step!

 

[gopher]

Ok i thought i got Now but i guess i dont.....

\(\displaystyle \displaystyle\int \displaystyle\frac{4x}{x^2+4}\,dx\)
i got

\(\displaystyle \displaystyle\int 4x\displaystyle\frac{1}{x^2+4}\,dx\)

\(\displaystyle 4\displaystyle\int x\displaystyle\frac{1}{x^2+4}\,dx\)

\(\displaystyle =4\displaystyle\frac{1}{2}x^2 \ln |x^2+4| + C\)

\(\displaystyle =2x^2 \ln |x^2+4| + C\)

But my calc says the answer is

\(\displaystyle 2 * \ln (x^2+4)\)

Do i have to use the substitution rule here?
Is there a site online that will show the work so i can check my self ?

 

[skeeter]

this is quite a conceptual "leap" for someone who

... finally started integration in class.

\(\displaystyle \L \int \frac{4x}{x^2+4} dx =2\int \frac{2x}{x^2+4} dx\)

note that the integrand now has the form \(\displaystyle \L \frac{u'}{u}\), whose antiderivative is \(\displaystyle \L \ln|u| + C\) ...

\(\displaystyle \L 2\ln{(x^2+4)} + C\)

 

[ChaoticLlama]

Ok i thought i got Now but i guess i dont.....

\(\displaystyle \displaystyle\int \displaystyle\frac{4x}{x^2+4}\,dx

=2x^2 \ln |x^2+4| + C\)

But my calc says the answer is


Do i have to use the substitution rule here?
Is there a site online that will show the work so i can check my self ?

gopher, I have a feeling you didn't understand my post.
They are not individual functions, and are not treated as such

Just differentiate your answer, and if you don't get what you started with, you did it wrong.