What is the integral of (tan(x))^{1/3}sec^4(x) ?

The integral of (tan(x))^{1/3}sec^4(x) is 3/4 tan^{4/3}(x)+3/10 tan^{10/3}(x)+C

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integral of (tan(x))^{1/3}sec^4(x)

\int (tan (x))^{\frac{1}{3}} sec^{4} (x) d x

\frac{3}{4} tan^{\frac{4}{3}} (x) + \frac{3}{10} tan^{\frac{10}{3}} (x) + C

Solution steps

\int (tan (x))^{\frac{1}{3}} sec^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u^{\frac{1}{3}} (1 + u^{2}) : u^{\frac{1}{3}} + u^{\frac{7}{3}}

= \int u^{\frac{1}{3}} + u^{\frac{7}{3}} 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{1}{3}} d u + \int u^{\frac{7}{3}} d u

\int u^{\frac{1}{3}} d u = \frac{3}{4} u^{\frac{4}{3}}

\int u^{\frac{7}{3}} d u = \frac{3}{10} u^{\frac{10}{3}}

= \frac{3}{4} u^{\frac{4}{3}} + \frac{3}{10} u^{\frac{10}{3}}

Substitute back

u = tan (x)

= \frac{3}{4} tan^{\frac{4}{3}} (x) + \frac{3}{10} tan^{\frac{10}{3}} (x)

Add a constant to the solution

= \frac{3}{4} tan^{\frac{4}{3}} (x) + \frac{3}{10} tan^{\frac{10}{3}} (x) + C

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What is the integral of (tan(x))^{1/3}sec^4(x) ?
The integral of (tan(x))^{1/3}sec^4(x) is 3/4 tan^{4/3}(x)+3/10 tan^{10/3}(x)+C