What is the integral of cos^8(3x)sin^3(3x) ?

The integral of cos^8(3x)sin^3(3x) is 1/3 (-(cos^9(3x))/9+(cos^{11}(3x))/(11))+C

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integral of cos^8(3x)sin^3(3x)

\int cos^{8} (3 x) sin^{3} (3 x) d x

\frac{1}{3} (- \frac{cos ^{9} ( 3 x )}{9} + \frac{cos ^{11} ( 3 x )}{11}) + C

Solution steps

\int cos^{8} (3 x) sin^{3} (3 x) d x

Apply u-substitution

: \int cos^{8} (u) sin^{3} (u) \frac{1}{3} d u

= \int cos^{8} (u) sin^{3} (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 cos^{8} (u) sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{8} (u) d u

Apply u-substitution

: \int - v^{8} (1 - v^{2}) d v

= \frac{1}{3} \cdot \int - v^{8} (1 - v^{2}) d v

Expand

- v^{8} (1 - v^{2}) : - v^{8} + v^{10}

= \frac{1}{3} \cdot \int - v^{8} + v^{10} 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 v^{8} d v + \int v^{10} d v)

\int v^{8} d v = \frac{v ^{9}}{9}

\int v^{10} d v = \frac{v ^{11}}{11}

= \frac{1}{3} (- \frac{v ^{9}}{9} + \frac{v ^{11}}{11})

Substitute back

= \frac{1}{3} (- \frac{cos ^{9} ( 3 x )}{9} + \frac{cos ^{11} ( 3 x )}{11})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{9} ( 3 x )}{9} + \frac{cos ^{11} ( 3 x )}{11}) + C

(3 x - 3 x^{2}) tan (x)

9 (t + 1) (\frac{d y}{d t}) - 6 y = 18 t

\frac{x ^{2} - 4 x}{( x - 2 ) ^{2}}

\int \frac{2 ^{(18 x)}}{4} d x

\int 2 x x d x

What is the integral of cos^8(3x)sin^3(3x) ?
The integral of cos^8(3x)sin^3(3x) is 1/3 (-(cos^9(3x))/9+(cos^{11}(3x))/(11))+C