Evaluate the integral of $ cos^3(x)sin(x) $ with respect to $ x $
Answer 1
Joseph Robinson
To evaluate the integral of $ \cos^3(x)\sin(x) $ with respect to $ x $, we use a substitution method:
Let $ u = \cos(x) $, then $ du = -\sin(x) dx $. Consequently:
$ \int \cos^3(x)\sin(x) dx = \int u^3 (-du) = -\int u^3 du $
Now integrate:
$ -\int u^3 du = -\frac{u^4}{4} + C $
Substitute back $ \cos(x) $ for $ u $:
$ -\frac{\cos^4(x)}{4} + C $
The final answer is:
$ -\frac{\cos^4(x)}{4} + C $
Answer 2
Maria Rodriguez
To integrate $ cos^3(x)sin(x) $, we use substitution:
Let $ u = cos(x) $. Then:
$ du = -sin(x) dx $
So:
$ int cos^3(x)sin(x) dx = -int u^3 du $
Integrate:
$ -frac{u^4}{4} + C $
Substitute back:
$ -frac{cos^4(x)}{4} + C $
Answer 3
William King
Use substitution:
Let $ u = cos(x) $. Then:
$ du = -sin(x) dx $
So:
$ int cos^3(x)sin(x) dx = -int u^3 du $
Integrate:
$ -frac{u^4}{4} + C $
Substitute back:
$ -frac{cos^4(x)}{4} + C $
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