What is the integral of sqrt(sec(x))tan^5(x) ?

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

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integral of sqrt(sec(x))tan^5(x)

\int sec (x) tan^{5} (x) d x

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

Solution steps

\int sec (x) tan^{5} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec (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 = sec (x)

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

Add a constant to the solution

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

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