What is the integral of 9sin^6(x)cos(x) ?

The integral of 9sin^6(x)cos(x) is 9/7 sin^7(x)+C

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integral of 9sin^6(x)cos(x)

\int 9 sin^{6} (x) cos (x) d x

\frac{9}{7} sin^{7} (x) + C

Solution steps

\int 9 sin^{6} (x) cos (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^{6} (x) cos (x) d x

Apply u-substitution

: \int u^{6} d u

= 9 \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 9 \cdot \frac{u ^{7}}{7}

Substitute back

u = sin (x)

= 9 \cdot \frac{sin ^{7} ( x )}{7}

Simplify

9 \cdot \frac{sin ^{7} ( x )}{7} : \frac{9}{7} sin^{7} (x)

= \frac{9}{7} sin^{7} (x)

Add a constant to the solution

= \frac{9}{7} sin^{7} (x) + C

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What is the integral of 9sin^6(x)cos(x) ?
The integral of 9sin^6(x)cos(x) is 9/7 sin^7(x)+C