What is the integral of 18cos^2(x)sin(x) ?

The integral of 18cos^2(x)sin(x) is -6cos^3(x)+C

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integral of 18cos^2(x)sin(x)

\int 18 cos^{2} (x) sin (x) d x

- 6 cos^{3} (x) + C

Solution steps

\int 18 cos^{2} (x) sin (x) d x

Take the constant out:

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

= 18 \cdot \int cos^{2} (x) sin (x) d x

Apply u-substitution

: \int - u^{2} d u

= 18 \cdot \int - u^{2} d u

Take the constant out:

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

= 18 (- \int u^{2} d u)

Apply the Power Rule

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

= 18 (- \frac{u ^{3}}{3})

Substitute back

u = cos (x)

= 18 (- \frac{cos ^{3} ( x )}{3})

Simplify

18 (- \frac{cos ^{3} ( x )}{3}) : - 6 cos^{3} (x)

= - 6 cos^{3} (x)

Add a constant to the solution

= - 6 cos^{3} (x) + C

x \to 0 lim (\frac{3 - 5 x}{4 x})

\int \frac{13}{x ^{2} + 1} d x

\frac{d y}{d x} = 3 y e^{x + 8}

x \to \infty lim ((- 1)^{x} \cdot (1 - \frac{1}{x ^{2}}))

\int sin^{2} (φ) d φ

What is the integral of 18cos^2(x)sin(x) ?
The integral of 18cos^2(x)sin(x) is -6cos^3(x)+C