What is the integral of sin^8(x)*cos^5(x) ?

The integral of sin^8(x)*cos^5(x) is (sin^9(x))/9-(2sin^{11}(x))/(11)+(sin^{13}(x))/(13)+C

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integral of sin^8(x)*cos^5(x)

\int sin^{8} (x) \cdot cos^{5} (x) d x

\frac{sin ^{9} ( x )}{9} - \frac{2 sin ^{11} ( x )}{11} + \frac{sin ^{13} ( x )}{13} + C

Solution steps

\int sin^{8} (x) cos^{5} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x))^{2} cos (x) sin^{8} (x) d x

Apply u-substitution

: \int u^{8} (1 - u^{2})^{2} d u

= \int u^{8} (1 - u^{2})^{2} d u

Expand

u^{8} (1 - u^{2})^{2} : u^{8} - 2 u^{10} + u^{12}

= \int u^{8} - 2 u^{10} + u^{12} 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^{8} d u - \int 2 u^{10} d u + \int u^{12} d u

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

\int 2 u^{10} d u = \frac{2 u ^{11}}{11}

\int u^{12} d u = \frac{u ^{13}}{13}

= \frac{u ^{9}}{9} - \frac{2 u ^{11}}{11} + \frac{u ^{13}}{13}

Substitute back

u = sin (x)

= \frac{sin ^{9} ( x )}{9} - \frac{2 sin ^{11} ( x )}{11} + \frac{sin ^{13} ( x )}{13}

Add a constant to the solution

= \frac{sin ^{9} ( x )}{9} - \frac{2 sin ^{11} ( x )}{11} + \frac{sin ^{13} ( x )}{13} + C

\frac{d}{d x} ((2 x + 8)^{2})

\int 2 x^{3} d x

(\frac{2}{t})^{^{'}}

\frac{1}{5 x + 7}

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

What is the integral of sin^8(x)*cos^5(x) ?
The integral of sin^8(x)*cos^5(x) is (sin^9(x))/9-(2sin^{11}(x))/(11)+(sin^{13}(x))/(13)+C