What is the integral of cos^5(9-11x) ?

The integral of cos^5(9-11x) is -1/11 (sin(9-11x)-(2sin^3(9-11x))/3+(sin^5(9-11x))/5)+C

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integral of cos^5(9-11x)

\int cos^{5} (9 - 11 x) d x

- \frac{1}{11} (sin (9 - 11 x) - \frac{2 sin ^{3} ( 9 - 11 x )}{3} + \frac{sin ^{5} ( 9 - 11 x )}{5}) + C

Solution steps

\int cos^{5} (9 - 11 x) d x

Apply u-substitution

: \int - \frac{1}{11} cos^{5} (u) d u

= \int - \frac{1}{11} cos^{5} (u) d u

Take the constant out:

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

= - \frac{1}{11} \cdot \int cos^{5} (u) d u

Rewrite using trig identities

= - \frac{1}{11} \cdot \int (1 - sin^{2} (u))^{2} cos (u) d u

Apply u-substitution

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

= - \frac{1}{11} \cdot \int (1 - v^{2})^{2} d v

Expand

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

= - \frac{1}{11} \cdot \int 1 - 2 v^{2} + v^{4} 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}{11} (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

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

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

= - \frac{1}{11} (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= - \frac{1}{11} (sin (9 - 11 x) - \frac{2 sin ^{3} ( 9 - 11 x )}{3} + \frac{sin ^{5} ( 9 - 11 x )}{5})

Add a constant to the solution

= - \frac{1}{11} (sin (9 - 11 x) - \frac{2 sin ^{3} ( 9 - 11 x )}{3} + \frac{sin ^{5} ( 9 - 11 x )}{5}) + C

\frac{d}{d x} (\frac{3 x - 4}{2 x - 8})

\int 8 x - 5 d x

\frac{1}{3 ( 1 + x ^{2} )}, \frac{1}{6} x^{2}

x y^{^{''}} + 2 y^{^{'}} = 4 x^{2}

\frac{d}{d x} (lo g_{4} (x^{2} + 3))

What is the integral of cos^5(9-11x) ?
The integral of cos^5(9-11x) is -1/11 (sin(9-11x)-(2sin^3(9-11x))/3+(sin^5(9-11x))/5)+C