What is the integral of sqrt(tan^3(x))sec^4(x) ?

The integral of sqrt(tan^3(x))sec^4(x) is 2/5 tan^{5/2}(x)+2/9 tan^{9/2}(x)+C

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integral of sqrt(tan^3(x))sec^4(x)

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

\frac{2}{5} tan^{\frac{5}{2}} (x) + \frac{2}{9} tan^{\frac{9}{2}} (x) + C

Solution steps

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

tan^{3} (x) = tan^{\frac{3}{2}} (x),

assuming

tan (x) \geq 0

= \int tan^{\frac{3}{2}} (x) sec^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int u^{\frac{3}{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{3}{2}} d u + \int u^{\frac{7}{2}} d u

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

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

Substitute back

u = tan (x)

= \frac{2}{5} tan^{\frac{5}{2}} (x) + \frac{2}{9} tan^{\frac{9}{2}} (x)

Add a constant to the solution

= \frac{2}{5} tan^{\frac{5}{2}} (x) + \frac{2}{9} tan^{\frac{9}{2}} (x) + C

\int x^{- 5} d x

\int \frac{2 x + 1}{x ^{2} + 6 x + 10} d x

t \to \infty lim (\frac{1}{4 t})

f (x) = \frac{x}{x + 6}

ln (x y)

What is the integral of sqrt(tan^3(x))sec^4(x) ?
The integral of sqrt(tan^3(x))sec^4(x) is 2/5 tan^{5/2}(x)+2/9 tan^{9/2}(x)+C