What is the integral of sin^3(t)cos^6(t) ?

The integral of sin^3(t)cos^6(t) is -(cos^7(t))/7+(cos^9(t))/9+C

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integral of sin^3(t)cos^6(t)

\int sin^{3} (t) cos^{6} (t) d t

- \frac{cos ^{7} ( t )}{7} + \frac{cos ^{9} ( t )}{9} + C

Solution steps

\int sin^{3} (t) cos^{6} (t) d t

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = cos (t)

= - \frac{cos ^{7} ( t )}{7} + \frac{cos ^{9} ( t )}{9}

Add a constant to the solution

= - \frac{cos ^{7} ( t )}{7} + \frac{cos ^{9} ( t )}{9} + C

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What is the integral of sin^3(t)cos^6(t) ?
The integral of sin^3(t)cos^6(t) is -(cos^7(t))/7+(cos^9(t))/9+C