What is the integral of 9sin^2(x)cos^3(x) ?

The integral of 9sin^2(x)cos^3(x) is 9((sin^3(x))/3-(sin^5(x))/5)+C

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integral of 9sin^2(x)cos^3(x)

\int 9 sin^{2} (x) cos^{3} (x) d x

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

Solution steps

\int 9 sin^{2} (x) cos^{3} (x) d x

Take the constant out:

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

= 9 \cdot \int sin^{2} (x) cos^{3} (x) d x

Rewrite using trig identities

= 9 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{2} (x) d x

Apply u-substitution

: \int u^{2} (1 - u^{2}) d u

= 9 \cdot \int u^{2} (1 - u^{2}) d u

Expand

u^{2} (1 - u^{2}) : u^{2} - u^{4}

= 9 \cdot \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

= 9 (\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}

= 9 (\frac{u ^{3}}{3} - \frac{u ^{5}}{5})

Substitute back

u = sin (x)

= 9 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5})

Add a constant to the solution

= 9 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

\frac{\partial}{\partial x} (cos (5 z) e^{3 x + 4 y} \cdot 3)

x \to 1 - lim (x ln (x))

y^{^{'}} - (2 x + 1) y + 3 x (5 x - 2) = 0

\frac{5}{x}, 1

\int_{2}^{7} \frac{1}{4 x + 8} d x

What is the integral of 9sin^2(x)cos^3(x) ?
The integral of 9sin^2(x)cos^3(x) is 9((sin^3(x))/3-(sin^5(x))/5)+C