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

Answer 1

See below.

#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#
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Answer 2

To find the integral of ( \frac{1}{2} + \sin(x) ) with respect to ( x ), you would integrate each term separately. The integral of ( \frac{1}{2} ) with respect to ( x ) is ( \frac{1}{2}x ), and the integral of ( \sin(x) ) with respect to ( x ) is ( -\cos(x) ). Therefore, the integral of ( \frac{1}{2} + \sin(x) ) with respect to ( x ) is ( \frac{1}{2}x - \cos(x) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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