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

The integral of tan^3(x)(sec(x))^{3/2} is 2/7 sec^{7/2}(x)-2/3 sec^{3/2}(x)+C

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

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

\frac{2}{7} sec^{\frac{7}{2}} (x) - \frac{2}{3} sec^{\frac{3}{2}} (x) + C

Solution steps

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

Simplify

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int u^{\frac{5}{2}} - u^{\frac{1}{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{5}{2}} d u - \int u^{\frac{1}{2}} d u

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

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

= \frac{2}{7} u^{\frac{7}{2}} - \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = sec (x)

= \frac{2}{7} sec^{\frac{7}{2}} (x) - \frac{2}{3} sec^{\frac{3}{2}} (x)

Add a constant to the solution

= \frac{2}{7} sec^{\frac{7}{2}} (x) - \frac{2}{3} sec^{\frac{3}{2}} (x) + C

\frac{\partial}{\partial y} (2 y^{2} + 3 x)

x \to 0 lim (\frac{3}{0})

n = 0 \sum \infty (\frac{1}{3})^{2 n}

\frac{5 x}{4 x ^{2} + 1}

\int \frac{x}{x ^{2} + 2 x + 2} d x

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