What is the integral of cos^5(x)sin^4(x) ?

The integral of cos^5(x)sin^4(x) is (sin^5(x))/5-(2sin^7(x))/7+(sin^9(x))/9+C

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integral of cos^5(x)sin^4(x)

\int cos^{5} (x) sin^{4} (x) d x

\frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9} + C

Solution steps

\int cos^{5} (x) sin^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

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

= \frac{u ^{5}}{5} - \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = sin (x)

= \frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9}

Add a constant to the solution

= \frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9} + C

\frac{d y}{d x} 0.4 x y = 9 x

\int \frac{2 x - 11}{x ^{2} ( x + 2 )} d x

y^{^{'}} - 2 y = 3 e^{x}

\int_{2}^{10} \frac{x ^{2} + 36}{12 x} d x

x^{^{'''}} - x^{^{''}} - x^{^{'}} + x = 2 e^{- t} + 3

What is the integral of cos^5(x)sin^4(x) ?
The integral of cos^5(x)sin^4(x) is (sin^5(x))/5-(2sin^7(x))/7+(sin^9(x))/9+C