What is the integral of sin^6(x)*cos^5(x) ?

The integral of sin^6(x)*cos^5(x) is (sin^7(x))/7-(2sin^9(x))/9+(sin^{11}(x))/(11)+C

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integral of sin^6(x)*cos^5(x)

\int sin^{6} (x) \cdot cos^{5} (x) d x

\frac{sin ^{7} ( x )}{7} - \frac{2 sin ^{9} ( x )}{9} + \frac{sin ^{11} ( x )}{11} + C

Solution steps

\int sin^{6} (x) cos^{5} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x))^{2} cos (x) sin^{6} (x) d x

Apply u-substitution

: \int u^{6} (1 - u^{2})^{2} d u

= \int u^{6} (1 - u^{2})^{2} d u

Expand

u^{6} (1 - u^{2})^{2} : u^{6} - 2 u^{8} + u^{10}

= \int u^{6} - 2 u^{8} + u^{10} 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^{6} d u - \int 2 u^{8} d u + \int u^{10} d u

\int u^{6} d u = \frac{u ^{7}}{7}

\int 2 u^{8} d u = \frac{2 u ^{9}}{9}

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

= \frac{u ^{7}}{7} - \frac{2 u ^{9}}{9} + \frac{u ^{11}}{11}

Substitute back

u = sin (x)

= \frac{sin ^{7} ( x )}{7} - \frac{2 sin ^{9} ( x )}{9} + \frac{sin ^{11} ( x )}{11}

Add a constant to the solution

= \frac{sin ^{7} ( x )}{7} - \frac{2 sin ^{9} ( x )}{9} + \frac{sin ^{11} ( x )}{11} + C

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What is the integral of sin^6(x)*cos^5(x) ?
The integral of sin^6(x)*cos^5(x) is (sin^7(x))/7-(2sin^9(x))/9+(sin^{11}(x))/(11)+C