What is the integral of sin^6(x/2)cos^5(x/2) ?

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

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integral of sin^6(x/2)cos^5(x/2)

\int sin^{6} (\frac{x}{2}) cos^{5} (\frac{x}{2}) d x

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

Solution steps

\int sin^{6} (\frac{x}{2}) cos^{5} (\frac{x}{2}) d x

Apply u-substitution

: \int sin^{6} (w) cos^{5} (w) \cdot 2 d w

= \int sin^{6} (w) cos^{5} (w) \cdot 2 d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sin^{6} (w) cos^{5} (w) d w

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= 2 \cdot \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

= 2 (\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}

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

Substitute back

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

Add a constant to the solution

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

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