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

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

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integral of (sin^5(θ))/(sqrt(cos(θ)))

\int \frac{sin ^{5} ( θ )}{cos ( θ )} d θ

- 2 cos (θ) + \frac{4}{5} cos^{\frac{5}{2}} (θ) - \frac{2}{9} cos^{\frac{9}{2}} (θ) + C

Solution steps

\int \frac{sin ^{5} ( θ )}{cos ( θ )} d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int - \frac{1}{u} + 2 u^{\frac{3}{2}} - u^{\frac{7}{2}} 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 \frac{1}{u} d u + \int 2 u^{\frac{3}{2}} d u - \int u^{\frac{7}{2}} d u

\int \frac{1}{u} d u = 2 u

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

= - 2 u + \frac{4}{5} u^{\frac{5}{2}} - \frac{2}{9} u^{\frac{9}{2}}

Substitute back

u = cos (θ)

= - 2 cos (θ) + \frac{4}{5} cos^{\frac{5}{2}} (θ) - \frac{2}{9} cos^{\frac{9}{2}} (θ)

Add a constant to the solution

= - 2 cos (θ) + \frac{4}{5} cos^{\frac{5}{2}} (θ) - \frac{2}{9} cos^{\frac{9}{2}} (θ) + C

f (x) = (3 x^{3} + 3 x) (x - 3) (x + 5)

e^{x} x^{4} + 8 e^{x} x^{3} + 12 e^{x} x^{2}

\frac{d y}{d x} = 1 - y + x^{2} - y x^{2}

\int_{0}^{1} (x^{2} + 1) e^{- x} d x

n = 0 \sum \infty \frac{2}{5 ^{n}}

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