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

The integral of cos^9(x) is sin(x)-(4sin^3(x))/3+(6sin^5(x))/5-(4sin^7(x))/7+(sin^9(x))/9+C

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integral of cos^9(x)

\int cos^{9} (x) d x

sin (x) - \frac{4 sin ^{3} ( x )}{3} + \frac{6 sin ^{5} ( x )}{5} - \frac{4 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9} + C

Solution steps

\int cos^{9} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x))^{4} cos (x) d x

Apply u-substitution

: \int (1 - u^{2})^{4} d u

= \int (1 - u^{2})^{4} d u

Expand

(1 - u^{2})^{4} : 1 - 4 u^{2} + 6 u^{4} - 4 u^{6} + u^{8}

= \int 1 - 4 u^{2} + 6 u^{4} - 4 u^{6} + u^{8} 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 4 u^{2} d u + \int 6 u^{4} d u - \int 4 u^{6} d u + \int u^{8} d u

\int 1 d u = u

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

\int 6 u^{4} d u = \frac{6 u ^{5}}{5}

\int 4 u^{6} d u = \frac{4 u ^{7}}{7}

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

= u - \frac{4 u ^{3}}{3} + \frac{6 u ^{5}}{5} - \frac{4 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = sin (x)

= sin (x) - \frac{4 sin ^{3} ( x )}{3} + \frac{6 sin ^{5} ( x )}{5} - \frac{4 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9}

Add a constant to the solution

= sin (x) - \frac{4 sin ^{3} ( x )}{3} + \frac{6 sin ^{5} ( x )}{5} - \frac{4 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9} + C

\int (8 x^{3} - 3 x^{2} + 8) d x

\frac{d}{d x} (x^{5} e^{6 x})

x \to \infty lim (\frac{2}{x})

\frac{d}{d x} (- 2 cos (3 x))

x \to 0 + lim (\frac{x}{3 x} - 7 x)

What is the integral of cos^9(x) ?
The integral of cos^9(x) is sin(x)-(4sin^3(x))/3+(6sin^5(x))/5-(4sin^7(x))/7+(sin^9(x))/9+C