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

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

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integral of sin^3(2x)

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

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

Solution steps

\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

f^{^{''}} (x) = x^{- \frac{3}{2}}

\int x 4 - x d x

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

y = \frac{a ^{4}}{a ^{3} - 8 x}

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