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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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

Add a constant to the solution

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

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