What is the integral of \sqrt[5]{tan(x)}sec^4(x) ?

The integral of \sqrt[5]{tan(x)}sec^4(x) is 5/6 tan^{6/5}(x)+5/16 tan^{16/5}(x)+C

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integral of \sqrt[5]{tan(x)}sec^4(x)

\frac{5}{6} tan^{\frac{6}{5}} (x) + \frac{5}{16} tan^{\frac{16}{5}} (x) + C

Solution steps

Rewrite using trig identities

Apply u-substitution

Expand

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{11}{5}} d u = \frac{5}{16} u^{\frac{16}{5}}

= \frac{5}{6} u^{\frac{6}{5}} + \frac{5}{16} u^{\frac{16}{5}}

Substitute back

u = tan (x)

= \frac{5}{6} tan^{\frac{6}{5}} (x) + \frac{5}{16} tan^{\frac{16}{5}} (x)

Add a constant to the solution

= \frac{5}{6} tan^{\frac{6}{5}} (x) + \frac{5}{16} tan^{\frac{16}{5}} (x) + C

What is the integral of \sqrt[5]{tan(x)}sec^4(x) ?
The integral of \sqrt[5]{tan(x)}sec^4(x) is 5/6 tan^{6/5}(x)+5/16 tan^{16/5}(x)+C