What is the integral of sin^3(θ) ?

The integral of sin^3(θ) is -cos(θ)+(cos^3(θ))/3+C

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integral of sin^3(θ)

\int sin^{3} (θ) d θ

- cos (θ) + \frac{cos ^{3} ( θ )}{3} + C

Solution steps

\int sin^{3} (θ) d θ

Rewrite using trig identities

= \int (1 - cos^{2} (θ)) sin (θ) d θ

Apply u-substitution

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

= \int - 1 + u^{2} 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

= - \int 1 d u + \int u^{2} d u

\int 1 d u = u

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

= - u + \frac{u ^{3}}{3}

Substitute back

u = cos (θ)

= - cos (θ) + \frac{cos ^{3} ( θ )}{3}

Add a constant to the solution

= - cos (θ) + \frac{cos ^{3} ( θ )}{3} + C

\frac{\partial}{\partial u} (\frac{u + v}{3})

49 y^{^{''}} + 126 y^{^{'}} + 81 y = 0

\int_{0}^{\frac{π}{3}} tan^{3} (x) d x

\int cos^{3} (\frac{x}{19}) d x

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