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

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

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integral of (cos^5(x))/(sqrt(sin(x)))

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

2 sin (x) - \frac{4}{5} sin^{\frac{5}{2}} (x) + \frac{2}{9} sin^{\frac{9}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

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 = sin (x)

= 2 sin (x) - \frac{4}{5} sin^{\frac{5}{2}} (x) + \frac{2}{9} sin^{\frac{9}{2}} (x)

Add a constant to the solution

= 2 sin (x) - \frac{4}{5} sin^{\frac{5}{2}} (x) + \frac{2}{9} sin^{\frac{9}{2}} (x) + C

n = 1 \sum \infty \frac{1}{n n}

\int_{\frac{3}{2}}^{\frac{5}{2}} e^{- s t} d t

y^{^{'}} = 4 - y

\int (\frac{2}{x ^{3}} + \frac{3}{x ^{2}} + 5) d x

\frac{e ^{- s}}{s ( s + 2 )}

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