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

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

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integral of sin^3(x)*cos^2(x)

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

- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5} + C

Solution steps

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

Rewrite using trig identities

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

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 = cos (x)

= - \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}

Add a constant to the solution

= - \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5} + C

\frac{\partial}{\partial x} (5 x^{4} y^{- 4})

\int a^{2 x} d x

\int arcsin (b x) d x

y \frac{d y}{d x} = sin (x)

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

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