What is the integral of tan^{5/3}(x)sec^6(x) ?

The integral of tan^{5/3}(x)sec^6(x) is 3/8 tan^{8/3}(x)+3/7 tan^{14/3}(x)+3/20 tan^{20/3}(x)+C

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integral of tan^{5/3}(x)sec^6(x)

\int tan^{\frac{5}{3}} (x) sec^{6} (x) d x

\frac{3}{8} tan^{\frac{8}{3}} (x) + \frac{3}{7} tan^{\frac{14}{3}} (x) + \frac{3}{20} tan^{\frac{20}{3}} (x) + C

Solution steps

\int tan^{\frac{5}{3}} (x) sec^{6} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{\frac{5}{3}} (x) d x

Apply u-substitution

: \int w^{\frac{5}{3}} (1 + w^{2})^{2} d w

= \int w^{\frac{5}{3}} (1 + w^{2})^{2} d w

Expand

w^{\frac{5}{3}} (1 + w^{2})^{2} : w^{\frac{5}{3}} + 2 w^{\frac{11}{3}} + w^{\frac{17}{3}}

= \int w^{\frac{5}{3}} + 2 w^{\frac{11}{3}} + w^{\frac{17}{3}} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int w^{\frac{5}{3}} d w + \int 2 w^{\frac{11}{3}} d w + \int w^{\frac{17}{3}} d w

\int w^{\frac{5}{3}} d w = \frac{3}{8} w^{\frac{8}{3}}

\int 2 w^{\frac{11}{3}} d w = \frac{3}{7} w^{\frac{14}{3}}

\int w^{\frac{17}{3}} d w = \frac{3}{20} w^{\frac{20}{3}}

= \frac{3}{8} w^{\frac{8}{3}} + \frac{3}{7} w^{\frac{14}{3}} + \frac{3}{20} w^{\frac{20}{3}}

Substitute back

w = tan (x)

= \frac{3}{8} tan^{\frac{8}{3}} (x) + \frac{3}{7} tan^{\frac{14}{3}} (x) + \frac{3}{20} tan^{\frac{20}{3}} (x)

Add a constant to the solution

= \frac{3}{8} tan^{\frac{8}{3}} (x) + \frac{3}{7} tan^{\frac{14}{3}} (x) + \frac{3}{20} tan^{\frac{20}{3}} (x) + C

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What is the integral of tan^{5/3}(x)sec^6(x) ?
The integral of tan^{5/3}(x)sec^6(x) is 3/8 tan^{8/3}(x)+3/7 tan^{14/3}(x)+3/20 tan^{20/3}(x)+C