Help? Integral of 2sec^3x dx

In order to calculate this integral, we need to use integration by parts.

By pulling out the 2:
∫ 2sec^3(x) dx = 2 ∫ sec^3(x) dx.

Next, integrate ∫ sec^3(x) dx by parts with:
u = sec(x) ==> du = sec(x)tan(x) dx
dv = sec^2(x) dx ==> v = tan(x).

This yields:
2 ∫ sec^3(x) dx = 2(uv - ∫ v du)
= 2sec(x)tan(x) - 2 ∫ sec(x)tan^2(x) dx
= 2sec(x)tan(x) - 2 ∫ sec(x)[sec^2(x) - 1] dx, since tan^2(x) = sec^2(x) - 1
= 2sec(x)tan(x) - 2 ∫ [sec^3(x) - sec(x)] dx, by expanding
= 2sec(x)tan(x) - 2 ∫ sec^3(x) dx + ∫ sec(x) dx
= 2sec(x)tan(x) + 2ln|sec(x) + tan(x)| - 2 ∫ sec^3(x) dx, by integrating ∫ sec(x) dx.

At this point, we have:
2 ∫ sec^3(x) dx = 2sec(x)tan(x) + 2ln|sec(x) + tan(x)| - 2 ∫ sec^3(x) dx.

Adding 2 ∫ sec^3(x) dx to both sides:
4 ∫ sec^3(x) dx = 2sec(x)tan(x) + 2ln|sec(x) + tan(x)|.

Dividing both sides by 2 and tacking on the integration constant:
∫ 2sec^3(x) dx = sec(x)tan(x) + ln|sec(x) + tan(x)| + C.

I hope this helps!

Evaluate the integral by parts:

∫ 2sec³x dx
2 ∫ sec³x dx
∫sec³x dx = ∫sec²xsecx dx
Let u = secx
du = secxtanx dx
dv = sec²x dx
v = tanx
∫udv = uv - ∫vdu
∫sec³x dx = secxtanx - ∫tan²xsecx dx
∫sec³x dx = secxtanx - ∫sec³x - secx dx
∫sec³x dx = secxtanx - ∫sec³x dx + ∫secx dx
2∫sec³x dx= secxtanx + ln|secx + tanx| + C


∫ 2sec³x dx = secxtanx + ln|secx + tanx| + C

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Let I = ∫2*sec^3(x) dx

Integrate by parts:

u = 2*sec(x)

dv = sec^2(x) dx

v = tan(x)

du = 2*sec(x)*tan(x) dx

I = 2*sec(x)*tan(x) - ∫2*sec(x)*tan^2(x) dx

I = 2*sec(x)*tan(x) - 2*∫sec(x)*(sec^2(x) - 1)) dx

I = 2*sec(x)*tan(x) - I + 2*ln|sec(x) + tan(x)| + C

I = sec(x)*tan(x) + ln|sec(x) + tan(x)| + C