Question

what is the integral of 1/2x+sinx?

Answer 1

\(\large \int\dfrac{1}{2x}+\sin x\;dx\)
Like this?

Or is the sine suppose to be in the denominator?

Answer 2

it is \[\int\limits(1\div2)x + sinx\]

Answer 3

my computer is silly and doesn't do the fraction properly

Answer 4

oh i see :)

Answer 5

\[\large \int\limits \frac{1}{2}x+\sin x\;dx\]We can integrate each term individually if that will help,\[\large =\int\limits \frac{1}{2}x\;dx+\int\limits\sin x\;dx\]

For the x term we just need to apply the `Power Rule for Integration`.
Remember how to do that?

It involves two steps:
~Add 1 to the power.
~Then divide by the new power.

\(\large \int x^n\;dx=\frac{1}{n+1}x^{n+1}\)

Answer 6

The 1/2 is a constant, let's pull that out of the integral so we don't have to worry about it.
\[\large \frac{1}{2}\int\limits x^1\;dx\]Our x is x to the `first` power.
Applying this rule gives us,\[\large \frac{1}{2}\left(\frac{1}{1+1}x^{1+1}\right)\qquad =\qquad \frac{1}{2}\left(\frac{1}{2}x^2\right)\]

Answer 7

Understand that part? :o

Answer 8

yep :) could you just make it 1/4 x^2?

Answer 9

Yes that sounds right.
So here's where we're at currently,
\[\large =\int\limits\limits \frac{1}{2}x\;dx+\int\limits\limits\sin x\;dx\]


\[\large =\frac{1}{4}x^2+\int\limits\limits\sin x\;dx\]

Answer 10

Do you remember your trig derivatives? :O

Answer 11

-cosx

Answer 12

\[\large \frac{1}{4}x^2-\cos x+C\]
Yay good job :D Psshh you didn't need help!
Seems like you've got this pretty good.
Just don't forget to add a constant of integration since it's an indefinite integral.

Answer 13

yep :) I just wasn't sure what to do because there were was more than 1 x in the equation

Answer 14

ah c:

Answer 15

thanks :P