What is the integral of cos^{10}(2y)sin^3(2y) ?

The integral of cos^{10}(2y)sin^3(2y) is 1/2 (-(cos^{11}(2y))/(11)+(cos^{13}(2y))/(13))+C

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integral of cos^{10}(2y)sin^3(2y)

\int cos^{10} (2 y) sin^{3} (2 y) d y

\frac{1}{2} (- \frac{cos ^{11} ( 2 y )}{11} + \frac{cos ^{13} ( 2 y )}{13}) + C

Solution steps

\int cos^{10} (2 y) sin^{3} (2 y) d y

Apply u-substitution

: \int cos^{10} (u) sin^{3} (u) \frac{1}{2} d u

= \int cos^{10} (u) sin^{3} (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 cos^{10} (u) sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{10} (u) d u

Apply u-substitution

: \int - v^{10} (1 - v^{2}) d v

= \frac{1}{2} \cdot \int - v^{10} (1 - v^{2}) d v

Expand

- v^{10} (1 - v^{2}) : - v^{10} + v^{12}

= \frac{1}{2} \cdot \int - v^{10} + v^{12} 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^{10} d v + \int v^{12} d v)

\int v^{10} d v = \frac{v ^{11}}{11}

\int v^{12} d v = \frac{v ^{13}}{13}

= \frac{1}{2} (- \frac{v ^{11}}{11} + \frac{v ^{13}}{13})

Substitute back

= \frac{1}{2} (- \frac{cos ^{11} ( 2 y )}{11} + \frac{cos ^{13} ( 2 y )}{13})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{11} ( 2 y )}{11} + \frac{cos ^{13} ( 2 y )}{13}) + C

\frac{d}{d x} (\frac{1}{9} x - ln (3 x))

y = x^{2}, y = 8 x

y^{^{'}} - 6 x^{2} y = 0, y (1) = e

y = \frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}}

What is the integral of cos^{10}(2y)sin^3(2y) ?
The integral of cos^{10}(2y)sin^3(2y) is 1/2 (-(cos^{11}(2y))/(11)+(cos^{13}(2y))/(13))+C