What is #int sinx cosx #?

Answer 1

#\intcos(x)sin(x)dx=\frac{-1}{4}cos(2x)+c#

Given equation is #\intcos(x)sin(x)dx#
I believe you're familiar with the trigonometric identity #sin(2\theta)=2sin(\theta)cos(\theta)# So, the equation becomes #\frac{1}{2}\intsin(2x)dx#

I'm sure you know what to do from here to get the equation I got there above.

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Answer 2

There are 3 equivalent results:
#(sin^2x)/2+C#
#-(cos^2x)/2+C#
#-(cos(2x))/4+C#

I found an awesome video about this:
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Answer 3

The integral of ( \sin(x) \cos(x) ) is ( -\frac{1}{2} \cos^2(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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