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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

= \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 = sin (t)

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

Add a constant to the solution

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

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