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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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

Add a constant to the solution

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

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