What is the integral of 5sin^2(r)cos(r) ?

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

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integral of 5sin^2(r)cos(r)

\int 5 sin^{2} (r) cos (r) d r

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

Solution steps

\int 5 sin^{2} (r) cos (r) d r

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int sin^{2} (r) cos (r) d r

Apply u-substitution

: \int u^{2} d u

= 5 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 5 \cdot \frac{u ^{3}}{3}

Substitute back

u = sin (r)

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

Simplify

5 \cdot \frac{sin ^{3} ( r )}{3} : \frac{5}{3} sin^{3} (r)

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

Add a constant to the solution

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

x^{2 x + 1}

x \to 5 lim (\frac{1}{x - 5})

\frac{d}{d x} ((ln (x))^{x})

(3, 12), (6, 13)

y = \frac{1}{x}

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