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

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

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integral of sin^2(θ)cos^3(θ)

\int sin^{2} (θ) cos^{3} (θ) d θ

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

Solution steps

\int sin^{2} (θ) cos^{3} (θ) d θ

Rewrite using trig identities

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

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

Add a constant to the solution

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

y = (10 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

\int \frac{x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\int (8 x^{4} + 5)^{3} 32 x^{3} d x

\frac{d}{d x} (\frac{7}{x + 2})

\frac{2 x + 1}{x + 2}, at a = 1

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