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

The integral of (sin(2x))^3(cos(2x))^2 is 1/2 (-(cos^3(2x))/3+(cos^5(2x))/5)+C

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

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

\frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3} + \frac{cos ^{5} ( 2 x )}{5}) + C

Solution steps

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

Simplify

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- v^{2} (1 - v^{2}) : - v^{2} + v^{4}

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

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{2} (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3} + \frac{cos ^{5} ( 2 x )}{5})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3} + \frac{cos ^{5} ( 2 x )}{5}) + C

\int (1 + x^{2}) e^{- x} d x

n = 0 \sum \infty \frac{n}{3 n ^{3} + 1}

\frac{d y}{d x} = \frac{( x + y )}{3 x + 3 y - 4}

(2 x + 3 y) d x + (3 x + 2 y) d y = 0

9 - 6 x

What is the integral of (sin(2x))^3(cos(2x))^2 ?
The integral of (sin(2x))^3(cos(2x))^2 is 1/2 (-(cos^3(2x))/3+(cos^5(2x))/5)+C