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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (t)

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

Add a constant to the solution

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

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