What is the integral of (1-sin^2(3t))cos(3t) ?

The integral of (1-sin^2(3t))cos(3t) is 1/3 (sin(3t)-(sin^3(3t))/3)+C

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integral of (1-sin^2(3t))cos(3t)

\int (1 - sin^{2} (3 t)) cos (3 t) d t

\frac{1}{3} (sin (3 t) - \frac{sin ^{3} ( 3 t )}{3}) + C

Solution steps

\int (1 - sin^{2} (3 t)) cos (3 t) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

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

Substitute back

= \frac{1}{3} (sin (3 t) - \frac{sin ^{3} ( 3 t )}{3})

Add a constant to the solution

= \frac{1}{3} (sin (3 t) - \frac{sin ^{3} ( 3 t )}{3}) + C

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What is the integral of (1-sin^2(3t))cos(3t) ?
The integral of (1-sin^2(3t))cos(3t) is 1/3 (sin(3t)-(sin^3(3t))/3)+C