How do you find the antiderivative of #int cos^3xsin^2xdx#?

Answer 1

#1/3sin^3x - 1/5sin^5x + C#

Factor:

#intcosx(cos^2x)sin^2xdx#
Rewrite using the pythagorean identity #sin^2theta + cos^2theta = 1#:
#intcosx(1 - sin^2x)sin^2xdx#

Expand:

#intcosx(sin^2x - sin^4x)dx#
Let #u = sinx#. Then #du = cosxdx = dx = (du)/cosx#.
#intcosx(u^2 - u^4) * (du)/cosx#
#intu^2 - u^4du#
This can be integrated as #intx^ndx = x^(n + 1)/(n +1) + C#, where #n != -1#
#1/3u^3 - 1/5u^5 + C#

Reverse the substitution:

#1/3sin^3x - 1/5sin^5x + C#

Hopefully this helps!

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Answer 2

To find the antiderivative of ∫cos^3(x)sin^2(x)dx, use the substitution method. Let u = sin(x), then du = cos(x)dx. After substitution, the integral becomes ∫u^2 du, which is straightforward to integrate. Finally, substitute back u = sin(x) to obtain the final antiderivative.

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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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