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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (r)

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

Add a constant to the solution

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

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