What's the integral of $\int tan x {(sec x)}^{\frac{3}{2}} d x$ $int tanx(secx)^(3/2) dx$ ?

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What's the integral of $\int tan x {(sec x)}^{\frac{3}{2}} d x$ $int tanx(secx)^(3/2) dx$ ?

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Answer 1 · 2015-11-12T08:52:35

$\int tan x {(sec x)}^{\frac{3}{2}} d x = \frac{2}{3} {(sec x)}^{\frac{3}{2}} + C$ $inttanx(secx)^(3/2)dx = 2/3(secx)^(3/2) + C$

Explanation:

For this integral, we will use $u$ $u$ substitution.
Let $u = sec x$ $u = secx$
then $d u = sec x tan x d x$ $du = secxtanxdx$

So we have
$\int tan x {(sec x)}^{\frac{3}{2}} d x = \int {(sec x)}^{\frac{1}{2}} \cdot sec x tan x d x = \int u^{\frac{1}{2}} d u$ $inttanx(secx)^(3/2)dx = int(secx)^(1/2)*secxtanxdx=intu^(1/2)du$

Applying the formula $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n+1)/(n+1) + C$ we get

$\int u^{\frac{1}{2}} d u = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{3} u^{\frac{3}{2}} + C$ $intu^(1/2)du = u^(3/2)/(3/2)+C = 2/3u^(3/2) + C$

Finally, we substitute back in for $u$ $u$ :

$\frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{3} {(sec x)}^{\frac{3}{2}} + C$ $2/3u^(3/2) + C = 2/3(secx)^(3/2) + C$

Thus

$\int tan x {(sec x)}^{\frac{3}{2}} d x = \frac{2}{3} {(sec x)}^{\frac{3}{2}} + C$ $inttanx(secx)^(3/2)dx = 2/3(secx)^(3/2) + C$

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What's the integral of $\int tan x {(sec x)}^{\frac{3}{2}} d x$ $int tanx(secx)^(3/2) dx$ ?

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