What is the integral of sqrt(tan(x))sec^6(x) ?

The integral of sqrt(tan(x))sec^6(x) is 2/3 tan^{3/2}(x)+4/7 tan^{7/2}(x)+2/11 tan^{11/2}(x)+C

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integral of sqrt(tan(x))sec^6(x)

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

\frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{4}{7} tan^{\frac{7}{2}} (x) + \frac{2}{11} tan^{\frac{11}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u (1 + u^{2})^{2} : u + 2 u^{\frac{5}{2}} + u^{\frac{9}{2}}

= \int u + 2 u^{\frac{5}{2}} + u^{\frac{9}{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 d u + \int 2 u^{\frac{5}{2}} d u + \int u^{\frac{9}{2}} d u

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

\int 2 u^{\frac{5}{2}} d u = \frac{4}{7} u^{\frac{7}{2}}

\int u^{\frac{9}{2}} d u = \frac{2}{11} u^{\frac{11}{2}}

= \frac{2}{3} u^{\frac{3}{2}} + \frac{4}{7} u^{\frac{7}{2}} + \frac{2}{11} u^{\frac{11}{2}}

Substitute back

u = tan (x)

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{4}{7} tan^{\frac{7}{2}} (x) + \frac{2}{11} tan^{\frac{11}{2}} (x)

Add a constant to the solution

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{4}{7} tan^{\frac{7}{2}} (x) + \frac{2}{11} tan^{\frac{11}{2}} (x) + C

y = (1 + 3 x)^{8}, (0, 1)

\int (x^{2} e^{2 x}) d x

x \to - \infty lim (f (x))

\frac{d}{d x} (4 x^{3} - 156 x^{2} + 1512 x)

\frac{d}{d x} (y e^{x^{2}})

What is the integral of sqrt(tan(x))sec^6(x) ?
The integral of sqrt(tan(x))sec^6(x) is 2/3 tan^{3/2}(x)+4/7 tan^{7/2}(x)+2/11 tan^{11/2}(x)+C