Compute the following trig integral

(sqrt sec(x)) tan^3(x) dx

We've been using the trig identity tan^2(x) +1 = sec^2(x), but I can't seem to solve it, even after I break apart tan^3(x) to be tan(x) and tan^2(x). Help?

sec^(1/2)(x) tan^3(x) dx

[sec(x)]^(1/2) sin^3(x) sec^3(x) dx

[sec(x)]^(7/2) sin^3(x) dx

Let u = sec(x)
du = sec(x) tan(x) dx = sec²(x) sin(x) dx

sin²(x) + cos²(x) = 1
sin²(x) = 1 - cos²(x)
sin²(x) = 1 - (1/sec²(x))

u^(3/2) sin^2(x) dx
u^(3/2) * [1 - (1/u²)] dx
[u^(3/2) - u^(-1/2)] dx
Integrate and substitute back.

Hello,

∫ [√(secx)] tan³x dx =

multiply and divide the integrand by √(secx):

∫ {[√(secx)√(secx)] /√(secx)} tan³x dx =

∫ [secx /√(secx)] tan³x dx =

rearrange this as:

∫ [tan²x /√(secx)] tanx secx dx =

replace tan²x with sec²x - 1:

∫ [(sec²x - 1) /√(secx)] tanx secx dx =

(distributing and simplifying)

∫ {[sec²x /√(secx)] - [1 /√(secx)]} tanx secx dx =

∫ {(secx)^[2 - (1/2)] - (secx)^(-1/2)} tanx secx dx =

∫ [(secx)^(3/2) - (secx)^(-1/2)] tanx secx dx =

let:

secx = u

differentiate both sides:

d(secx) = du → tanx secx dx = du

then, substituting:

∫ [(secx)^(3/2) - (secx)^(-1/2)] tanx secx dx = ∫ [u^(3/2) - u^(-1/2)] du =

(splitting into two integrals)

∫ u^(3/2) du - ∫ u^(-1/2) du =

{1/[(3/2)+1]} u^[(3/2)+1] - {1/[(-1/2)+1]} u^[(-1/2)+1] + C =

[1/(5/2)]u^(5/2) - [1/(1/2)]u^(1/2) + C =

(2/5)u^(5/2) - 2u^(1/2) + C =

(2/5)√u^5 - 2√u + C

substitute back for u, ending with:


∫ [√(secx)] tan³x dx = (2/5)√(secx)^5 - 2√(secx) + C



I hope it helps