What is the integral of sin^4(v)cos^3(v) ?

The integral of sin^4(v)cos^3(v) is (sin^5(v))/5-(sin^7(v))/7+C

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integral of sin^4(v)cos^3(v)

\int sin^{4} (v) cos^{3} (v) d v

\frac{sin ^{5} ( v )}{5} - \frac{sin ^{7} ( v )}{7} + C

Solution steps

\int sin^{4} (v) cos^{3} (v) d v

Rewrite using trig identities

= \int (1 - sin^{2} (v)) cos (v) sin^{4} (v) d v

Apply u-substitution

: \int u^{4} (1 - u^{2}) d u

= \int u^{4} (1 - u^{2}) d u

Expand

u^{4} (1 - u^{2}) : u^{4} - u^{6}

= \int u^{4} - u^{6} 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^{4} d u - \int u^{6} d u

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

\int u^{6} d u = \frac{u ^{7}}{7}

= \frac{u ^{5}}{5} - \frac{u ^{7}}{7}

Substitute back

u = sin (v)

= \frac{sin ^{5} ( v )}{5} - \frac{sin ^{7} ( v )}{7}

Add a constant to the solution

= \frac{sin ^{5} ( v )}{5} - \frac{sin ^{7} ( v )}{7} + C

x y^{^{'}} + y = 2 x y^{2}, y (1) = - 5

\frac{\partial}{\partial x} (e^{y} sin (x y))

\frac{d}{d x} (\frac{12 x ^{2} - x ^{4}}{( 4 - x ^{2} ) ^{2}})

f (x) = (x - 3) (2 x + 4)

ln (x^{3} - 2 x)

What is the integral of sin^4(v)cos^3(v) ?
The integral of sin^4(v)cos^3(v) is (sin^5(v))/5-(sin^7(v))/7+C