What is the integral of sin^3(19x) ?

The integral of sin^3(19x) is 1/19 (-cos(19x)+(cos^3(19x))/3)+C

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integral of sin^3(19x)

\int sin^{3} (19 x) d x

\frac{1}{19} (- cos (19 x) + \frac{cos ^{3} ( 19 x )}{3}) + C

Solution steps

\int sin^{3} (19 x) d x

Apply u-substitution

: \int sin^{3} (u) \frac{1}{19} d u

= \int sin^{3} (u) \frac{1}{19} d u

Take the constant out:

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

= \frac{1}{19} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{19} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= \frac{1}{19} \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}{19} (- \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}{19} (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{19} (- cos (19 x) + \frac{cos ^{3} ( 19 x )}{3})

Add a constant to the solution

= \frac{1}{19} (- cos (19 x) + \frac{cos ^{3} ( 19 x )}{3}) + C

n = 5 \sum \infty \frac{n}{( n + 1 ) !}

\int \frac{8 sec ^{2} ( 4 t )}{3 + 2 tan ( 4 t )} d t

\frac{x}{x ^{2} + 16}

\int \frac{x ^{2} - 4}{x - 2} d x

f (x) = (3 x^{2} + 2)^{2} (x^{2} - 5 x)^{3}

What is the integral of sin^3(19x) ?
The integral of sin^3(19x) is 1/19 (-cos(19x)+(cos^3(19x))/3)+C