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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

= \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 = cos (r)

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

Add a constant to the solution

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

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