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

The integral of sin(wt)cos^2(wt) is -1/(3w)cos^3(wt)+C

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integral of sin(wt)cos^2(wt)

\int sin (wt) cos^{2} (wt) d t

- \frac{1}{3 w} cos^{3} (wt) + C

Solution steps

\int sin (wt) cos^{2} (wt) d t

Apply u-substitution

: \int \frac{cos ^{2} ( u ) sin ( u )}{w} d u

= \int \frac{cos ^{2} ( u ) sin ( u )}{w} d u

Take the constant out:

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

= \frac{1}{w} \cdot \int cos^{2} (u) sin (u) d u

Apply u-substitution

: \int - v^{2} d v

= \frac{1}{w} \cdot \int - v^{2} d v

Take the constant out:

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

= \frac{1}{w} (- \int v^{2} d v)

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{w} (- \frac{v ^{3}}{3})

Substitute back

= \frac{1}{w} (- \frac{cos ^{3} ( wt )}{3})

Simplify

\frac{1}{w} (- \frac{cos ^{3} ( wt )}{3}) : - \frac{1}{3 w} cos^{3} (wt)

= - \frac{1}{3 w} cos^{3} (wt)

Add a constant to the solution

= - \frac{1}{3 w} cos^{3} (wt) + C

x^{3} - 3 x + 2

\int (\frac{e}{3})^{3 x} d x

\frac{d ^{3}}{d x ^{3}} (\frac{9}{x ^{2}})

\int + sin^{3} (θ) cos^{8} (θ) d θ

\frac{d}{d x} (\frac{15}{x ^{4}})

What is the integral of sin(wt)cos^2(wt) ?
The integral of sin(wt)cos^2(wt) is -1/(3w)cos^3(wt)+C