Question

How do you find the integral of `1/2+sinx dx`?

Answer 1

See below.

Explanation:

`int(1/2+sinx)dx`

It is a known result that the derivative of `sinx` is `cosx`, and that the derivative of `cosx` is `-sinx`. It should therefore follow that the derivative of `-cosx` is `sinx` (by just swapping the signs of the previous one).
As a result, `intsinxdx=-cosx` since it is just the reverse.

Furthermore, since `intf(x)+-g(x)dx=intf(x)dx+-intg(x)dx`
Then : `int(1/2+sinx)dx=int1/2dx+intsinxdx`

By employing the integral value for algebraic functions of x and our result from earlier, this becomes equal to `x/2+A-cosx+B` (`A` and `B` are the constants produced by integration).

Finally, we can add together the constants and have our final answer as:

`x/2-cosx+C`