What is the integral of sin^3(5x) ?

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

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integral of sin^3(5x)

\int sin^{3} (5 x) d x

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

Solution steps

\int sin^{3} (5 x) d x

Apply u-substitution

: \int sin^{3} (u) \frac{1}{5} d u

= \int sin^{3} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

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

Substitute back

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

Add a constant to the solution

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

\int \frac{1}{x ( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

\int_{0}^{20} 4 + 3 x d x

y^{^{''}} - 3 y^{^{'}} - 4 y = - 3, y (0) = 0, y^{^{'}} (0) = 0

\int tan^{2} (2 x se) c^{2} 2 x

y^{^{'}} + y - 2 cos (x) = 0

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