What is the integral of sin^3(2x)cos^7(2x) ?

The integral of sin^3(2x)cos^7(2x) is 1/2 (-(cos^8(2x))/8+(cos^{10}(2x))/(10))+C

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integral of sin^3(2x)cos^7(2x)

\int sin^{3} (2 x) cos^{7} (2 x) d x

\frac{1}{2} (- \frac{cos ^{8} ( 2 x )}{8} + \frac{cos ^{10} ( 2 x )}{10}) + C

Solution steps

\int sin^{3} (2 x) cos^{7} (2 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{7} (u) \frac{1}{2} d u

= \int sin^{3} (u) cos^{7} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin^{3} (u) cos^{7} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{7} (u) d u

Apply u-substitution

: \int - v^{7} (1 - v^{2}) d v

= \frac{1}{2} \cdot \int - v^{7} (1 - v^{2}) d v

Expand

- v^{7} (1 - v^{2}) : - v^{7} + v^{9}

= \frac{1}{2} \cdot \int - v^{7} + v^{9} 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}{2} (- \int v^{7} d v + \int v^{9} d v)

\int v^{7} d v = \frac{v ^{8}}{8}

\int v^{9} d v = \frac{v ^{10}}{10}

= \frac{1}{2} (- \frac{v ^{8}}{8} + \frac{v ^{10}}{10})

Substitute back

= \frac{1}{2} (- \frac{cos ^{8} ( 2 x )}{8} + \frac{cos ^{10} ( 2 x )}{10})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{8} ( 2 x )}{8} + \frac{cos ^{10} ( 2 x )}{10}) + C

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What is the integral of sin^3(2x)cos^7(2x) ?
The integral of sin^3(2x)cos^7(2x) is 1/2 (-(cos^8(2x))/8+(cos^{10}(2x))/(10))+C