What is the integral of (cos^3(θ))(sin^5(θ)) ?

The integral of (cos^3(θ))(sin^5(θ)) is -(cos^4(θ))/4+(cos^6(θ))/3-(cos^8(θ))/8+C

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integral of (cos^3(θ))(sin^5(θ))

\int (cos^{3} (θ)) (sin^{5} (θ)) d θ

- \frac{cos ^{4} ( θ )}{4} + \frac{cos ^{6} ( θ )}{3} - \frac{cos ^{8} ( θ )}{8} + C

Solution steps

\int cos^{3} (θ) sin^{5} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{3} (1 - u^{2})^{2} : - u^{3} + 2 u^{5} - u^{7}

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

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

\int 2 u^{5} d u = \frac{u ^{6}}{3}

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

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{3} - \frac{u ^{8}}{8}

Substitute back

u = cos (θ)

= - \frac{cos ^{4} ( θ )}{4} + \frac{cos ^{6} ( θ )}{3} - \frac{cos ^{8} ( θ )}{8}

Add a constant to the solution

= - \frac{cos ^{4} ( θ )}{4} + \frac{cos ^{6} ( θ )}{3} - \frac{cos ^{8} ( θ )}{8} + C

\int \frac{e ^{- x}}{x ^{4}} d x

\int (cos^{6} (x)) d x

e^{- x^{3}}

\int (\frac{2}{t} - \frac{3}{t ^{2}}) d t

\int \frac{4 ^{c o t (5 x)}}{sin ^{2} ( 5 x )} d x

What is the integral of (cos^3(θ))(sin^5(θ)) ?
The integral of (cos^3(θ))(sin^5(θ)) is -(cos^4(θ))/4+(cos^6(θ))/3-(cos^8(θ))/8+C