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

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

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

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

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

Solution steps

\int sin^{4} (x) cos^{5} (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}{d x} (x (1 + x)^{- 1})

f (z) = (9 z^{4} + 3 z^{2} + 1) (3 z^{3} - z)

\int (4 - 2 x)^{2} d x

x \to - 8 lim (9 x + 10)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = \frac{e ^{x}}{x}

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