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

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

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integral of sin^3(θ)cos^2(θ)

\int sin^{3} (θ) cos^{2} (θ) d θ

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

Solution steps

\int sin^{3} (θ) cos^{2} (θ) d θ

Rewrite using trig identities

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

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 (θ)

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

Add a constant to the solution

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

\int 6 e^{4 x} cos (8 x) d x

y = \frac{1}{x - 2}, at x = 6

\frac{d}{d x} ((\frac{4}{5} x)^{5})

\int \frac{x}{( x ^{4} + 2 )} d x

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 7

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