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

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

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integral of sin^9(x)

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

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

Solution steps

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

Rewrite using trig identities

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

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

Add a constant to the solution

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

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