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

The integral of (sin(2x))^3 is 8(-(cos^4(x))/4+(cos^6(x))/6)+C

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integral of (sin(2x))^3

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

8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}) + C

Solution steps

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

Simplify

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

Rewrite using trig identities

= \int 8 cos^{3} (x) sin^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int cos^{3} (x) sin^{3} (x) d x

Rewrite using trig identities

= 8 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{3} (x) d x

Apply u-substitution

: \int - u^{3} (1 - u^{2}) d u

= 8 \cdot \int - u^{3} (1 - u^{2}) d u

Expand

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= 8 \cdot \int - u^{3} + u^{5} d u

Apply the Sum Rule:

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

= 8 (- \int u^{3} d u + \int u^{5} d u)

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{5} d u = \frac{u ^{6}}{6}

= 8 (- \frac{u ^{4}}{4} + \frac{u ^{6}}{6})

Substitute back

u = cos (x)

= 8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6})

Add a constant to the solution

= 8 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}) + C

\int (\frac{1}{x ^{3} + 1}) d x

x (2 x + 1)^{3}

\frac{d}{d x} (sinh^{2} (3 x) + arcsin (2 x))

\int (3 x^{3} + (2 x - 3)^{2}) d x

\frac{d}{d x} (tan (x) + cot (x))

What is the integral of (sin(2x))^3 ?
The integral of (sin(2x))^3 is 8(-(cos^4(x))/4+(cos^6(x))/6)+C