What is the integral of sin^8(r)cos^3(r) ?

The integral of sin^8(r)cos^3(r) is (sin^9(r))/9-(sin^{11}(r))/(11)+C

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integral of sin^8(r)cos^3(r)

\int sin^{8} (r) cos^{3} (r) d r

\frac{sin ^{9} ( r )}{9} - \frac{sin ^{11} ( r )}{11} + C

Solution steps

\int sin^{8} (r) cos^{3} (r) d r

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

= \int 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^{8} d u - \int u^{10} d u

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

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

= \frac{u ^{9}}{9} - \frac{u ^{11}}{11}

Substitute back

u = sin (r)

= \frac{sin ^{9} ( r )}{9} - \frac{sin ^{11} ( r )}{11}

Add a constant to the solution

= \frac{sin ^{9} ( r )}{9} - \frac{sin ^{11} ( r )}{11} + C

What is the integral of sin^8(r)cos^3(r) ?
The integral of sin^8(r)cos^3(r) is (sin^9(r))/9-(sin^{11}(r))/(11)+C