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

The integral of sin^5(x)cos^9(x) is -(cos^{10}(x)}{10}+(cos^{12}(x))/6-\frac{cos^{14}(x))/(14)+C

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integral of sin^5(x)cos^9(x)

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

- \frac{cos ^{10} ( x )}{10} + \frac{cos ^{12} ( x )}{6} - \frac{cos ^{14} ( x )}{14} + C

Solution steps

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

Rewrite using trig identities

= \int (1 - cos^{2} (x))^{2} sin (x) cos^{9} (x) d x

Apply u-substitution

: \int - u^{9} (1 - u^{2})^{2} d u

= \int - u^{9} (1 - u^{2})^{2} d u

Expand

- u^{9} (1 - u^{2})^{2} : - u^{9} + 2 u^{11} - u^{13}

= \int - u^{9} + 2 u^{11} - u^{13} 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 u^{9} d u + \int 2 u^{11} d u - \int u^{13} d u

\int u^{9} d u = \frac{u ^{10}}{10}

\int 2 u^{11} d u = \frac{u ^{12}}{6}

\int u^{13} d u = \frac{u ^{14}}{14}

= - \frac{u ^{10}}{10} + \frac{u ^{12}}{6} - \frac{u ^{14}}{14}

Substitute back

u = cos (x)

= - \frac{cos ^{10} ( x )}{10} + \frac{cos ^{12} ( x )}{6} - \frac{cos ^{14} ( x )}{14}

Add a constant to the solution

= - \frac{cos ^{10} ( x )}{10} + \frac{cos ^{12} ( x )}{6} - \frac{cos ^{14} ( x )}{14} + C

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What is the integral of sin^5(x)cos^9(x) ?
The integral of sin^5(x)cos^9(x) is -(cos^{10}(x)}{10}+(cos^{12}(x))/6-\frac{cos^{14}(x))/(14)+C