How do you integrate #sec^3x "d"x#?

Question

Elijah Abram · Accepted Answer

Perform integration by parts, then a substitution.

#int sec^3(x) dx = 1/2(tan(x)sec(x)+lnabs(secx+tanx))+ C# Since #sec^3(x) = sec^2(x) sec(x)#, #int sec^3(x)dx = int sec^2(x) sec(x)dx# With #u=sec(x) <=> u'=sec(x)tan(x)#, #v'=sec^2(x) <=> v=tan(x)#, and #int uv'dx=uv-int u'vdx#, we have #int sec^3(x)dx=tan(x)sec(x)-int sec(x)tan^2(x)dx# Since #sec^2(x)-1=tan^2(x)#, #int sec(x)tan^2(x)dx# #= int sec(x)(sec^2(x)-1)dx# #=int sec^3(x)dx-int sec(x)dx# Hence #int sec^3(x)dx# #=tan(x)sec(x)-(int sec^3(x)dx-int sec(x)dx)# #=tan(x)sec(x)-int sec^3(x)dx+int sec(x)dx# If we add #int sec^3(x)dx# to both sides, we have #2int sec^3(x)dx=tan(x)sec(x)+int sec(x)dx# #int sec^3(x)dx=1/2(tan(x)sec(x)+int sec(x)dx)# Now to figure out what #int sec(x)dx# is, you can either look it up in a formula sheet or derive it, as I will now. Now there are a couple ways to derive this, but I will use the shortest and most common method for this. #int sec(x)dx=int sec(x)/1dx# Now if I multiply the numerator and denominator by #sec x+tan x#, #int sec(x)dx=int (sec x(sec x+tan x))/(sec x + tan x)dx# #=int (sec^2(x)+secxtanx)/(sec x+ tan x)dx# Now let #u=sec x+tan x#, thus #du=(secxtanx+sec^2(x))dx# #int (sec^2(x)+secxtanx)/(sec x+ tan x)dx# #=int 1/u * (sec^2(x)+secxtanx)dx# #=int 1/u du = lnabs(u)+C=lnabs(secx+tanx)+C# Hence, given #int sec(x)dx=lnabs(secx+tanx)+C#, #int sec^3(x)dx=1/2(tan(x)sec(x)+int sec(x)dx)# #=1/2(tan(x)sec(x)+lnabs(secx+tanx)) + C#

Cameron Alexander · Answer

#intsec^3x"d"x=1/2(secxtanx+lnabs(secx+tanx))+"c"#

For #intsec^3x"d"x#, use integration by parts.

Let #u=secx# and #"d"u=secxtanx"d"x#

and #"d"v=sec^2x"d"x# thus #v=tanx#

We now enter this into the calculation.

#intu("d"v)/("d"x)"d"x=uv-intv("d"u)/("d"x)"d"x#

So

#intsec^3x"d"x=secxtanx-intsecxtan^2x"d"x#

#=secxtanx-intsecx(sec^2x-1)"d"x#

#=secxtanx-intsec^3x"d"x+intsecx"d"x#

Thus

#2intsec^3x"d"x=secxtanx+intsecx"d"x#

So

#2intsec^3x"d"x=secxtanx+lnabs(secx+tanx)+"C"#

So

#intsec^3"d"x=1/2(secxtanx+lnabs(secx+tanx))+"c"#

Christopher Allers · Answer

undefined To integrate sec^3(x) with respect to x, you can use integration by parts method or a substitution method. Here's the step-by-step process using integration by parts:

1. Let u = sec(x) and dv = sec^2(x)dx
2. Find du and v by differentiating and integrating u and dv respectively.
3. Apply the integration by parts formula: ∫u dv = uv - ∫v du
4. Substitute the values of u, v, du, and dv into the formula and solve the integral.

The integral of sec^3(x)dx will be:

∫sec^3(x)dx = sec(x)tan(x) - ∫sec(x)tan^2(x)dx

Now, you need to integrate ∫sec(x)tan^2(x)dx. This can be solved by using a trigonometric identity or substitution method. The result will involve sec(x) and tan(x) terms.