What is the integral of tan^5(θ)sec^{9/2}(θ) ?

The integral of tan^5(θ)sec^{9/2}(θ) is 2/17 sec^{17/2}(θ)-4/13 sec^{13/2}(θ)+2/9 sec^{9/2}(θ)+C

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integral of tan^5(θ)sec^{9/2}(θ)

\int tan^{5} (θ) sec^{\frac{9}{2}} (θ) d θ

\frac{2}{17} sec^{\frac{17}{2}} (θ) - \frac{4}{13} sec^{\frac{13}{2}} (θ) + \frac{2}{9} sec^{\frac{9}{2}} (θ) + C

Solution steps

\int tan^{5} (θ) sec^{\frac{9}{2}} (θ) d θ

Rewrite using trig identities

= \int (- 1 + sec^{2} (θ))^{2} tan (θ) sec^{\frac{9}{2}} (θ) d θ

Apply u-substitution

: \int u^{\frac{7}{2}} (u^{2} - 1)^{2} d u

= \int u^{\frac{7}{2}} (u^{2} - 1)^{2} d u

Expand

u^{\frac{7}{2}} (u^{2} - 1)^{2} : u^{\frac{15}{2}} - 2 u^{\frac{11}{2}} + u^{\frac{7}{2}}

= \int u^{\frac{15}{2}} - 2 u^{\frac{11}{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 u^{\frac{15}{2}} d u - \int 2 u^{\frac{11}{2}} d u + \int u^{\frac{7}{2}} d u

\int u^{\frac{15}{2}} d u = \frac{2}{17} u^{\frac{17}{2}}

\int 2 u^{\frac{11}{2}} d u = \frac{4}{13} u^{\frac{13}{2}}

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

= \frac{2}{17} u^{\frac{17}{2}} - \frac{4}{13} u^{\frac{13}{2}} + \frac{2}{9} u^{\frac{9}{2}}

Substitute back

u = sec (θ)

= \frac{2}{17} sec^{\frac{17}{2}} (θ) - \frac{4}{13} sec^{\frac{13}{2}} (θ) + \frac{2}{9} sec^{\frac{9}{2}} (θ)

Add a constant to the solution

= \frac{2}{17} sec^{\frac{17}{2}} (θ) - \frac{4}{13} sec^{\frac{13}{2}} (θ) + \frac{2}{9} sec^{\frac{9}{2}} (θ) + C

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What is the integral of tan^5(θ)sec^{9/2}(θ) ?
The integral of tan^5(θ)sec^{9/2}(θ) is 2/17 sec^{17/2}(θ)-4/13 sec^{13/2}(θ)+2/9 sec^{9/2}(θ)+C