How do you find the integral of #int cos x * sin x dx#?

Answer 1

It is

#int cosx*sinxdx=1/2 int sin2xdx=-1/4cos2x+c#
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Answer 2

To find the integral of ∫cos(x) * sin(x) dx, you can use integration by parts. Let u = cos(x) and dv = sin(x) dx. Then, differentiate u to get du and integrate dv to get v. After that, apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, dv, du, and v into the formula and solve the integral. This yields:

∫cos(x) * sin(x) dx = -cos(x) * cos(x) - ∫(-cos(x) * (-sin(x))) dx

Now, integrate the remaining term:

∫(-cos(x) * (-sin(x))) dx = ∫cos(x) * sin(x) dx

So, you get:

∫cos(x) * sin(x) dx = -cos^2(x) - ∫cos(x) * sin(x) dx

Rearrange the equation to solve for the integral:

2∫cos(x) * sin(x) dx = -cos^2(x)

Now, solve for the integral:

∫cos(x) * sin(x) dx = -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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