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How do you find the integral of #sin^2x cosx #?

Answer 1

Emma Aldridge

# int sin^2xcosxdx = 1/3sin^3x + C#

Let # I = int sin^2xcosxdx #

We can integrate this by substitution:

Let # u=sinx => (du)/dx=cosx # # => int ... cosxdx = int ... du#

And so, we can rewrite the integral as follows: # I = int (sin^2x)cosxdx # # I = int u^2du # # I = 1/3u^3 #

Ad re-substituting for #u # we gte: # I = 1/3sin^3x + C#

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Answer 2

Emma Aldridge

To find the integral of sin^2(x) cos(x), you can use the substitution method. Let u = sin(x), then du = cos(x) dx. Now, rewrite the integral in terms of u:

∫sin^2(x) cos(x) dx = ∫u^2 du

Now integrate ∫u^2 du, which gives:

= (1/3)u^3 + C

Substitute back u = sin(x):

= (1/3)sin^3(x) + C

So, the integral of sin^2(x) cos(x) is (1/3)sin^3(x) + 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.