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How do you find the indefinite integral of #int xsin3x^2dx#?

Answer 1

Emma Aldridge

I got: #-1/6cos(3x^2)+c#

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Answer 2

Ezra Adams

To find the indefinite integral of ( \int x \sin(3x^2) , dx ), you can use the substitution method. Let ( u = 3x^2 ), then ( du = 6x , dx ). This implies ( \frac{1}{6}du = x , dx ). Now, substituting these into the integral, we get ( \frac{1}{6} \int \sin(u) , du ). Integrating ( \sin(u) ) with respect to ( u ) yields ( -\frac{1}{6} \cos(u) + C ), where ( C ) is the constant of integration. Finally, substituting back ( u = 3x^2 ), the result is ( -\frac{1}{6} \cos(3x^2) + C ). Therefore, the indefinite integral of ( \int x \sin(3x^2) , dx ) is ( -\frac{1}{6} \cos(3x^2) + C ).

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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.