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

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

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

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

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

Solution steps

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

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

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

Simplify

sec (x) sec^{0.5} (x) : sec^{1.5} (x)

= \int sec^{1.5} (x) tan^{3} (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

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