What is the integral of tan^5(x)sec^{-3/2}(x) ?

The integral of tan^5(x)sec^{-3/2}(x) is -2/(3sec^{3/2)(x)}-4sec^{1/2}(x)+2/5 sec^{5/2}(x)+C

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integral of tan^5(x)sec^{-3/2}(x)

\int tan^{5} (x) sec^{- \frac{3}{2}} (x) d x

- \frac{2}{3 sec ^{\frac{3}{2}} ( x )} - 4 sec^{\frac{1}{2}} (x) + \frac{2}{5} sec^{\frac{5}{2}} (x) + C

Solution steps

\int tan^{5} (x) sec^{- \frac{3}{2}} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{- \frac{3}{2}} (x) d x

Apply u-substitution

: \int \frac{( - 1 + u ^{2} ) ^{2}}{u ^{\frac{5}{2}}} d u

= \int \frac{( - 1 + u ^{2} ) ^{2}}{u ^{\frac{5}{2}}} d u

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{5}{2}}} - \frac{2}{u ^{\frac{1}{2}}} + u^{\frac{3}{2}}

= \int \frac{1}{u ^{\frac{5}{2}}} - \frac{2}{u ^{\frac{1}{2}}} + u^{\frac{3}{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 ^{\frac{5}{2}}} d u - \int \frac{2}{u ^{\frac{1}{2}}} d u + \int u^{\frac{3}{2}} d u

\int \frac{1}{u ^{\frac{5}{2}}} d u = - \frac{2}{3 u ^{\frac{3}{2}}}

\int \frac{2}{u ^{\frac{1}{2}}} d u = 4 u^{\frac{1}{2}}

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

= - \frac{2}{3 u ^{\frac{3}{2}}} - 4 u^{\frac{1}{2}} + \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = sec (x)

= - \frac{2}{3 sec ^{\frac{3}{2}} ( x )} - 4 sec^{\frac{1}{2}} (x) + \frac{2}{5} sec^{\frac{5}{2}} (x)

Add a constant to the solution

= - \frac{2}{3 sec ^{\frac{3}{2}} ( x )} - 4 sec^{\frac{1}{2}} (x) + \frac{2}{5} sec^{\frac{5}{2}} (x) + C

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What is the integral of tan^5(x)sec^{-3/2}(x) ?
The integral of tan^5(x)sec^{-3/2}(x) is -2/(3sec^{3/2)(x)}-4sec^{1/2}(x)+2/5 sec^{5/2}(x)+C