What is the integral of cos^2(t)sin^5(t) ?

The integral of cos^2(t)sin^5(t) is -(cos^3(t))/3+(2cos^5(t))/5-(cos^7(t))/7+C

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integral of cos^2(t)sin^5(t)

\int cos^{2} (t) sin^{5} (t) d t

- \frac{cos ^{3} ( t )}{3} + \frac{2 cos ^{5} ( t )}{5} - \frac{cos ^{7} ( t )}{7} + C

Solution steps

\int cos^{2} (t) sin^{5} (t) d t

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{2} (1 - u^{2})^{2} : - u^{2} + 2 u^{4} - u^{6}

= \int - u^{2} + 2 u^{4} - u^{6} 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^{2} d u + \int 2 u^{4} d u - \int u^{6} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

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

= - \frac{u ^{3}}{3} + \frac{2 u ^{5}}{5} - \frac{u ^{7}}{7}

Substitute back

u = cos (t)

= - \frac{cos ^{3} ( t )}{3} + \frac{2 cos ^{5} ( t )}{5} - \frac{cos ^{7} ( t )}{7}

Add a constant to the solution

= - \frac{cos ^{3} ( t )}{3} + \frac{2 cos ^{5} ( t )}{5} - \frac{cos ^{7} ( t )}{7} + C

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What is the integral of cos^2(t)sin^5(t) ?
The integral of cos^2(t)sin^5(t) is -(cos^3(t))/3+(2cos^5(t))/5-(cos^7(t))/7+C