What is the integral of cos^5(θ)sin^4(θ) ?

The integral of cos^5(θ)sin^4(θ) is (sin^5(θ))/5-(2sin^7(θ))/7+(sin^9(θ))/9+C

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integral of cos^5(θ)sin^4(θ)

\int cos^{5} (θ) sin^{4} (θ) d θ

\frac{sin ^{5} ( θ )}{5} - \frac{2 sin ^{7} ( θ )}{7} + \frac{sin ^{9} ( θ )}{9} + C

Solution steps

\int cos^{5} (θ) sin^{4} (θ) d θ

Rewrite using trig identities

= \int (1 - sin^{2} (θ))^{2} cos (θ) sin^{4} (θ) d θ

Apply u-substitution

: \int u^{4} (1 - u^{2})^{2} d u

= \int u^{4} (1 - u^{2})^{2} d u

Expand

u^{4} (1 - u^{2})^{2} : u^{4} - 2 u^{6} + u^{8}

= \int u^{4} - 2 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^{4} d u - \int 2 u^{6} d u + \int u^{8} d u

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

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

= \frac{u ^{5}}{5} - \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = sin (θ)

= \frac{sin ^{5} ( θ )}{5} - \frac{2 sin ^{7} ( θ )}{7} + \frac{sin ^{9} ( θ )}{9}

Add a constant to the solution

= \frac{sin ^{5} ( θ )}{5} - \frac{2 sin ^{7} ( θ )}{7} + \frac{sin ^{9} ( θ )}{9} + C

(- 5, 6), (- 3, 4)

\int \frac{x ^{2}}{( 121 + x ^{2} ) ^{2}} d x

\frac{d}{d x} (\frac{x ^{4}}{2} - 3 x^{2} + 10)

\int \frac{y ^{3} e ^{y^{2}}}{( y ^{2} + 1 ) ^{2}} d y

f (x) = x^{2} - 2 x, (0, 2)

What is the integral of cos^5(θ)sin^4(θ) ?
The integral of cos^5(θ)sin^4(θ) is (sin^5(θ))/5-(2sin^7(θ))/7+(sin^9(θ))/9+C