What is the integral of 1/4 sin^3(x)cos(x) ?

The integral of 1/4 sin^3(x)cos(x) is 1/16 sin^4(x)+C

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integral of 1/4 sin^3(x)cos(x)

\int \frac{1}{4} sin^{3} (x) cos (x) d x

\frac{1}{16} sin^{4} (x) + C

Solution steps

\int \frac{1}{4} sin^{3} (x) cos (x) d x

Take the constant out:

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

= \frac{1}{4} \cdot \int sin^{3} (x) cos (x) d x

Apply u-substitution

: \int u^{3} d u

= \frac{1}{4} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{4} \cdot \frac{u ^{4}}{4}

Substitute back

u = sin (x)

= \frac{1}{4} \cdot \frac{sin ^{4} ( x )}{4}

Simplify

\frac{1}{4} \cdot \frac{sin ^{4} ( x )}{4} : \frac{1}{16} sin^{4} (x)

= \frac{1}{16} sin^{4} (x)

Add a constant to the solution

= \frac{1}{16} sin^{4} (x) + C

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What is the integral of 1/4 sin^3(x)cos(x) ?
The integral of 1/4 sin^3(x)cos(x) is 1/16 sin^4(x)+C