How do you find the integral of #tan^2(x) * sec^3(x) dx#?

Question

Luke Alvarez · Accepted Answer

See the explanation section, below.

Rewrite the integrand using #tan^2x = sec^2x-1#. Let's give the integral we want the name #I# #I = int tan^2xsec^3x dx = int (sec^5x-sec^3x)dx# Next we'll integrate #sec^5x# by parts. #int sec^5x dx = int sec^3 x sec^2x dx# Let #u = sec^3 x# and #dv = sec^2x dx#. Then #du = 3tanx sec^3x dx# and #v = tanx# We get #int sec^5 x dx = sec^3x tanx - 3int tan^2x sec^3x dx# Again, use #tan^2x = sec^2 x-1# to get #int sec^5 x dx = sec^3x tanx - 3int (sec^2 x-1) sec^3x dx# #int sec^5 x dx = sec^3x tanx - 3int sec^5 dx + 3int sec^3x dx# which brings us to #4int sec^5 x dx = sec^3x tanx + 3int sec^3x dx# and #int sec^5 x dx = 1/4sec^3x tanx + 3/4 int sec^3x dx# Recalling that the other integral we need is #int sec^3x dx#, let's simplify our lives by writing: #I = intsec^5xdx-intsec^3xdx# # = underbrace(1/4sec^3x tanx + 3/4 int sec^3x dx)_(intsec^5 x dx)-intsec^3xdx# # = 1/4sec^3x tanx -1/4 int sec^3x dx# So now find #int sec^3x dx = int secx sec^2x dx# using the same general approach and integration by parts. You should get #int sec^3x dx = 1/2secxtanx - 1/2int secx dx# Thus, as of right now, we have #I = 1/4sec^3x tanx -1/4underbrace( [1/2secxtanx - 1/2int secx dx])_(intsec^3 x dx)# # = 1/4sec^3x tanx -1/8secxtanx + 1/8 int secx dx# Now #int secx dx# can be evaluated in a couple of ways, the usual trick is to multiply by #(secx+tanx)/(secx+tanx)# to get #int 1/u du = ln absu = ln abs(secx+tanx)# Thus, we will now conclude with #I = 1/4sec^3x tanx - 1/8secxtanx + 1/8 ln abs(sec x + tan x ) +C# Alternative form for #int secx dx# #intsecx dx# can also be found by substitution and partial fractions to get #1/2ln abs((sinx+1)/(sinx-1))+C# (Yes, the is equivalent to # ln abs(secx+tanx)+C#)

Ezra Adams · Answer

undefined To integrate $ \tan^2(x) \cdot \sec^3(x) $ with respect to $ x $, you can use integration by parts. Let $ u = \tan(x) $ and $ dv = \sec^3(x) dx $. Then, $ du = \sec^2(x) dx $ and $ v = \frac{1}{2}\tan(x)\sec(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)| $.

Applying the integration by parts formula $ \int u \, dv = uv - \int v \, du $, we get:

$$ \int \tan^2(x) \cdot \sec^3(x) \, dx = \frac{1}{2}\tan(x)\sec(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)| - \int \sec^2(x) \cdot \left(\frac{1}{2}\tan(x)\sec(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)|\right) \, dx $$

$$ = \frac{1}{2}\tan(x)\sec(x) + \frac{1}{2}\ln|\sec(x) + \tan(x)| - \frac{1}{2} \int \tan(x)\sec^3(x) \, dx - \frac{1}{2} \int \sec(x)\ln|\sec(x) + \tan(x)| \, dx $$

The integral $ \int \tan(x)\sec^3(x) \, dx $ can be evaluated using substitution, and $ \int \sec(x)\ln|\sec(x) + \tan(x)| \, dx $ can be evaluated using integration by parts again.

This process may need to be repeated until you get integrals that you can directly evaluate. Once you've evaluated those integrals, you can substitute the results back into the original equation to find the final answer.