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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

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

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

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

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

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

Substitute back

u = sin (v)

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

Add a constant to the solution

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

\frac{d}{d x} ((sin (x)) (cos (x)))

\frac{d y}{d x}, x y = 15

\frac{\partial}{\partial x} (2 sin (x^{2} + y^{2}))

f (x) = (5)^{x}, at x = 4

\int x^{2} 1 - x^{3} d x

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