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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (t)

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

Add a constant to the solution

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

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