What is the integral of sin^3(x)cos^9(x) ?

The integral of sin^3(x)cos^9(x) is -(cos^{10}(x))/(10)+(cos^{12}(x))/(12)+C

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integral of sin^3(x)cos^9(x)

\int sin^{3} (x) cos^{9} (x) d x

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

Solution steps

\int sin^{3} (x) cos^{9} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = cos (x)

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{x}{y - x})

\frac{\partial}{\partial x} (x + yz)

\int_{1}^{2} \frac{13}{x ^{3} + 2 x} d x

\int \frac{2 x + 1}{x} d x

\frac{\partial}{\partial x} (e^{x y} cos (x^{2}))

What is the integral of sin^3(x)cos^9(x) ?
The integral of sin^3(x)cos^9(x) is -(cos^{10}(x))/(10)+(cos^{12}(x))/(12)+C