What is the integral of sin^3(y) ?

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

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integral of sin^3(y)

\int sin^{3} (y) d y

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

Solution steps

\int sin^{3} (y) d y

Rewrite using trig identities

= \int (1 - cos^{2} (y)) sin (y) d y

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 (y)

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

Add a constant to the solution

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

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