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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{8} e^{4 x})

\frac{d}{d x} (cos (0))

(3 x - 2)^{2} (x^{4} + 2)^{2}

\int \frac{2}{3} (1 - x)^{\frac{3}{2}} (1) d x

\frac{d}{d x} (\frac{6 x}{5 - tan ( x )})

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