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

1 Answer
May 20, 2016

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#