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) 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} (3 x - sin (x))

\int \frac{1}{7 - 7 x} d x

\int \frac{x ^{2} + x + 2}{x ^{4} + 5 x ^{2} + 4} d x

\int 6 x^{5} - 8 x^{4} - 9 x^{2} d 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