What is the integral of sin^5(t)cos^7(t) ?

The integral of sin^5(t)cos^7(t) is -(cos^8(t))/8+(cos^{10}(t))/5-(cos^{12}(t))/(12)+C

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integral of sin^5(t)cos^7(t)

\int sin^{5} (t) cos^{7} (t) d t

- \frac{cos ^{8} ( t )}{8} + \frac{cos ^{10} ( t )}{5} - \frac{cos ^{12} ( t )}{12} + C

Solution steps

\int sin^{5} (t) cos^{7} (t) d t

Rewrite using trig identities

= \int (1 - cos^{2} (t))^{2} sin (t) cos^{7} (t) d t

Apply u-substitution

: \int - u^{7} (1 - u^{2})^{2} d u

= \int - u^{7} (1 - u^{2})^{2} d u

Expand

- u^{7} (1 - u^{2})^{2} : - u^{7} + 2 u^{9} - u^{11}

= \int - u^{7} + 2 u^{9} - u^{11} 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^{7} d u + \int 2 u^{9} d u - \int u^{11} d u

\int u^{7} d u = \frac{u ^{8}}{8}

\int 2 u^{9} d u = \frac{u ^{10}}{5}

\int u^{11} d u = \frac{u ^{12}}{12}

= - \frac{u ^{8}}{8} + \frac{u ^{10}}{5} - \frac{u ^{12}}{12}

Substitute back

u = cos (t)

= - \frac{cos ^{8} ( t )}{8} + \frac{cos ^{10} ( t )}{5} - \frac{cos ^{12} ( t )}{12}

Add a constant to the solution

= - \frac{cos ^{8} ( t )}{8} + \frac{cos ^{10} ( t )}{5} - \frac{cos ^{12} ( t )}{12} + C

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What is the integral of sin^5(t)cos^7(t) ?
The integral of sin^5(t)cos^7(t) is -(cos^8(t))/8+(cos^{10}(t))/5-(cos^{12}(t))/(12)+C