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

The integral of cos^3(t)sin^3(t) is -(cos^4(t))/4+(cos^6(t))/6+C

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

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

- \frac{cos ^{4} ( t )}{4} + \frac{cos ^{6} ( t )}{6} + C

Solution steps

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

Rewrite using trig identities

= \int (1 - cos^{2} (t)) sin (t) cos^{3} (t) d t

Apply u-substitution

= \int - u^{3} (1 - u^{2}) d u

Expand

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= \int - u^{3} + u^{5} 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^{3} d u + \int u^{5} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{5} d u = \frac{u ^{6}}{6}

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

Substitute back

u = cos (t)

= - \frac{cos ^{4} ( t )}{4} + \frac{cos ^{6} ( t )}{6}

Add a constant to the solution

= - \frac{cos ^{4} ( t )}{4} + \frac{cos ^{6} ( t )}{6} + C

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What is the integral of cos^3(t)sin^3(t) ?
The integral of cos^3(t)sin^3(t) is -(cos^4(t))/4+(cos^6(t))/6+C