What is the integral of sin^3(φ) ?

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

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integral of sin^3(φ)

\int sin^{3} (φ) d φ

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

Solution steps

\int sin^{3} (φ) d φ

Rewrite using trig identities

= \int (1 - cos^{2} (φ)) sin (φ) d φ

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 (φ)

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

Add a constant to the solution

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

\int_{- e}^{e} \frac{1}{∣ x ∣} d x

1 + 2 x

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

f (x) = \frac{5 + csc ( x )}{9 - csc ( x )}

g

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