What is the integral of sin^3(16x) ?

The integral of sin^3(16x) is 1/16 (-cos(16x)+(cos^3(16x))/3)+C

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integral of sin^3(16x)

\int sin^{3} (16 x) d x

\frac{1}{16} (- cos (16 x) + \frac{cos ^{3} ( 16 x )}{3}) + C

Solution steps

\int sin^{3} (16 x) d x

Apply u-substitution

: \int sin^{3} (u) \frac{1}{16} d u

= \int sin^{3} (u) \frac{1}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{16} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

: \int - 1 + v^{2} d v

= \frac{1}{16} \cdot \int - 1 + v^{2} 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}{16} (- \int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= \frac{1}{16} (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{16} (- cos (16 x) + \frac{cos ^{3} ( 16 x )}{3})

Add a constant to the solution

= \frac{1}{16} (- cos (16 x) + \frac{cos ^{3} ( 16 x )}{3}) + C

f (x) = \frac{1}{( x ^{3} + 7 ) ^{4}}

\frac{d}{d x} (2 x^{3} - 4 x)

\frac{2}{3 x ^{4}}

n = 1 \sum \infty \frac{n}{5 + n ^{2}}

2 x - x^{2}, - 2 x

What is the integral of sin^3(16x) ?
The integral of sin^3(16x) is 1/16 (-cos(16x)+(cos^3(16x))/3)+C