How do you find the following indefinite integral of #(x^(2)+sin(2x))dx#?

Answer 1

I found: #x^3/3+sin^2(x)+c#

Consider: #int(x^2+sin(2x))dx=# #intx^2dx+intsin(2x)dx=# #=x^3/3+int2sin(x)cos(x)dx=# but #d[sin(x)]=cos(x)dx#; #x^3/3+2intsin(x)d[sin(x)]=x^3/3+cancel(2)sin^2(x)/cancel(2)+c#

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Answer 2

Isaiah Allen

To find the indefinite integral of ( \int (x^2 + \sin(2x)) , dx ), you can integrate each term separately. The integral of ( x^2 ) is ( \frac{x^3}{3} ) and the integral of ( \sin(2x) ) is ( -\frac{1}{2}\cos(2x) ). So, the indefinite integral of ( (x^2 + \sin(2x)) ) is ( \frac{x^3}{3} - \frac{1}{2}\cos(2x) + 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.