What is the integral of sin^3(4x)cos^6(4x) ?

The integral of sin^3(4x)cos^6(4x) is 1/4 (-(cos^7(4x))/7+(cos^9(4x))/9)+C

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integral of sin^3(4x)cos^6(4x)

\int sin^{3} (4 x) cos^{6} (4 x) d x

\frac{1}{4} (- \frac{cos ^{7} ( 4 x )}{7} + \frac{cos ^{9} ( 4 x )}{9}) + C

Solution steps

\int sin^{3} (4 x) cos^{6} (4 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{6} (u) \frac{1}{4} d u

= \int sin^{3} (u) cos^{6} (u) \frac{1}{4} d u

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} (u) cos^{6} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{6} (u) d u

Apply u-substitution

: \int - v^{6} (1 - v^{2}) d v

= \frac{1}{4} \cdot \int - v^{6} (1 - v^{2}) d v

Expand

- v^{6} (1 - v^{2}) : - v^{6} + v^{8}

= \frac{1}{4} \cdot \int - v^{6} + v^{8} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (- \int v^{6} d v + \int v^{8} d v)

\int v^{6} d v = \frac{v ^{7}}{7}

\int v^{8} d v = \frac{v ^{9}}{9}

= \frac{1}{4} (- \frac{v ^{7}}{7} + \frac{v ^{9}}{9})

Substitute back

= \frac{1}{4} (- \frac{cos ^{7} ( 4 x )}{7} + \frac{cos ^{9} ( 4 x )}{9})

Add a constant to the solution

= \frac{1}{4} (- \frac{cos ^{7} ( 4 x )}{7} + \frac{cos ^{9} ( 4 x )}{9}) + C

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What is the integral of sin^3(4x)cos^6(4x) ?
The integral of sin^3(4x)cos^6(4x) is 1/4 (-(cos^7(4x))/7+(cos^9(4x))/9)+C