What is the integral of tan^3(x)sec^{1/2}(x) ?

The integral of tan^3(x)sec^{1/2}(x) is -2sqrt(sec(x))+2/5 sec^{5/2}(x)+C

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integral of tan^3(x)sec^{1/2}(x)

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

- 2 sec (x) + \frac{2}{5} sec^{\frac{5}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{u} d u

= \int \frac{- 1 + u ^{2}}{u} d u

Expand

\frac{- 1 + u ^{2}}{u} : - \frac{1}{u} + u^{\frac{3}{2}}

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

\int \frac{1}{u} d u = 2 u

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

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

Substitute back

u = sec (x)

= - 2 sec (x) + \frac{2}{5} sec^{\frac{5}{2}} (x)

Add a constant to the solution

= - 2 sec (x) + \frac{2}{5} sec^{\frac{5}{2}} (x) + C

x \to 0 lim (\frac{1}{e ^{x}})

x \to 6 lim (\frac{x}{x ^{2} - 36})

\frac{d x}{d t} + x = - e^{t}

\frac{d}{d x} ((3 t^{2} - t + a^{2})^{101})

x \to + 3 + lim (\frac{7}{x - 3})

What is the integral of tan^3(x)sec^{1/2}(x) ?
The integral of tan^3(x)sec^{1/2}(x) is -2sqrt(sec(x))+2/5 sec^{5/2}(x)+C