What is the integral of (sin(3x))^7 ?

The integral of (sin(3x))^7 is -1/3 cos(3x)+1/3 cos^3(3x)-1/5 cos^5(3x)+1/21 cos^7(3x)+C

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integral of (sin(3x))^7

\int (sin (3 x))^{7} d x

- \frac{1}{3} cos (3 x) + \frac{1}{3} cos^{3} (3 x) - \frac{1}{5} cos^{5} (3 x) + \frac{1}{21} cos^{7} (3 x) + C

Solution steps

\int (sin (3 x))^{7} d x

Simplify

= \int sin^{7} (3 x) d x

Apply u-substitution

: \int sin^{7} (u) \frac{1}{3} d u

= \int sin^{7} (u) \frac{1}{3} 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 sin^{7} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - (1 - v^{2})^{3} d v

= \frac{1}{3} \cdot \int - (1 - v^{2})^{3} d v

Expand

- (1 - v^{2})^{3} : - 1 + 3 v^{2} - 3 v^{4} + v^{6}

= \frac{1}{3} \cdot \int - 1 + 3 v^{2} - 3 v^{4} + v^{6} 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 3 v^{2} d v - \int 3 v^{4} d v + \int v^{6} d v)

\int 1 d v = v

\int 3 v^{2} d v = v^{3}

\int 3 v^{4} d v = \frac{3 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{3} (- v + v^{3} - \frac{3 v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{3} (- cos (3 x) + cos^{3} (3 x) - \frac{3 cos ^{5} ( 3 x )}{5} + \frac{cos ^{7} ( 3 x )}{7})

Simplify

\frac{1}{3} (- cos (3 x) + cos^{3} (3 x) - \frac{3 cos ^{5} ( 3 x )}{5} + \frac{cos ^{7} ( 3 x )}{7}) : - \frac{1}{3} cos (3 x) + \frac{1}{3} cos^{3} (3 x) - \frac{1}{5} cos^{5} (3 x) + \frac{1}{21} cos^{7} (3 x)

= - \frac{1}{3} cos (3 x) + \frac{1}{3} cos^{3} (3 x) - \frac{1}{5} cos^{5} (3 x) + \frac{1}{21} cos^{7} (3 x)

Add a constant to the solution

= - \frac{1}{3} cos (3 x) + \frac{1}{3} cos^{3} (3 x) - \frac{1}{5} cos^{5} (3 x) + \frac{1}{21} cos^{7} (3 x) + C

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What is the integral of (sin(3x))^7 ?
The integral of (sin(3x))^7 is -1/3 cos(3x)+1/3 cos^3(3x)-1/5 cos^5(3x)+1/21 cos^7(3x)+C