What is the integral of cos^3(u)sin^2(u) ?

The integral of cos^3(u)sin^2(u) is (sin^3(u))/3-(sin^5(u))/5+C

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integral of cos^3(u)sin^2(u)

\int cos^{3} (u) sin^{2} (u) d u

\frac{sin ^{3} ( u )}{3} - \frac{sin ^{5} ( u )}{5} + C

Solution steps

\int cos^{3} (u) sin^{2} (u) d u

Rewrite using trig identities

= \int (1 - sin^{2} (u)) cos (u) sin^{2} (u) d u

Apply u-substitution

: \int v^{2} (1 - v^{2}) d v

= \int v^{2} (1 - v^{2}) d v

Expand

v^{2} (1 - v^{2}) : v^{2} - v^{4}

= \int v^{2} - v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int v^{2} d v - \int v^{4} d v

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{v ^{3}}{3} - \frac{v ^{5}}{5}

Substitute back

v = sin (u)

= \frac{sin ^{3} ( u )}{3} - \frac{sin ^{5} ( u )}{5}

Add a constant to the solution

= \frac{sin ^{3} ( u )}{3} - \frac{sin ^{5} ( u )}{5} + C

\int \frac{x ^{2} + 1}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (- \frac{3}{x ^{2}})

\frac{d y}{d x}, y = x^{- 10 c o s (x)}

f (x) = \frac{x ^{2} - 1}{x ^{2} + x + 1}, (1, 0)

(1 + x) \frac{d y}{d x} - x y = x + x^{2}

What is the integral of cos^3(u)sin^2(u) ?
The integral of cos^3(u)sin^2(u) is (sin^3(u))/3-(sin^5(u))/5+C