How do you integrate #int [(Sec(x))^5]dx#?

Question

Scarlett Allison · Accepted Answer

#int sec^5x dx = (2tanxsec^3x+ 3tanxsecx + 3ln abs(secx+tanx))/8 +C#

Write the integrand as: #sec^5(x) = sec^2(x) sec^3(x)# and integrate by parts considering that:

#d/dx (tanx) = sec^2(x) #,

so:

#int sec^5x dx = int sec^2(x) sec^3(x)dx#

#int sec^5x dx = int sec^3(x)d(tanx)#

#int sec^5x dx = tanxsec^3x - int tanx d(sec^3(x))#

and as:

#d/dx (sec^3(x)) = 3sec^2(x) d/dx sec(x) = 3sec^3(x) tanx#

we have:

#int sec^5x dx = tanxsec^3x - 3int tan^2x sec^3x dx#

use now the trigonometric identity:

#tan^2 theta = sin^2 theta/cos^2 theta = (1-cos^2 theta)/cos^2theta = sec^2theta -1#

to have:

#int sec^5x dx = tanxsec^3x - 3int (sec^2x -1) sec^3x dx#

and using the linearity of the integral:

#int sec^5x dx = tanxsec^3x + 3int sec^3x dx -3 int sec^5x dx#

The integral now appears on both sides of the equation and we can solve for it obtaining a reduction formula:

#int sec^5x dx = 1/4(tanxsec^3x + 3int sec^3x dx)#

Solve now the resulting integral with the same procedure:

#int sec^3x dx = int secx d(tanx)#

#int sec^3x dx = tanxsecx - int tanx d(secx)#

#int sec^3x dx = tanxsecx - int tan^2x secx dx#

#int sec^3x dx = tanxsecx - int (sec^2x-1) secx dx#

#int sec^3x dx = tanxsecx + int secx dx - int sec^3x dx#

#int sec^3x dx = 1/2(tanxsecx + int secx dx)#

To solve the resulting integral note that:

#d/dx (tanx + secx) = sec^2x +secx tanx = secx(tanx+secx)#

so dividing and multiplying the integrand by #(secx+tanx)#:

#int secx dx = int (secx(secx+tanx))/(secx+tanx) dx#

#int secx dx = int (d(secx+tanx))/(secx+tanx)#

#int secx dx = ln abs(secx+tanx) +C#

Putting it all together:

#int sec^5x dx = (2tanxsec^3x+ 3tanxsecx + 3ln abs(secx+tanx))/8 +C#

Elijah Abram · Answer

undefined To integrate $\int (\sec(x))^5 \, dx$, we can use the method of substitution. Here's the step-by-step process:

1. Let:
   $$ u = \sec(x) $$
   $$ du = \sec(x) \tan(x) \, dx $$

2. Rewrite the integral in terms of $u$:
   $$ \int (\sec(x))^5 \, dx = \int u^5 \, du $$

3. Now we integrate $u^5$ with respect to $u$:
   $$ \int u^5 \, du = \frac{u^6}{6} + C $$
   $$ = \frac{(\sec(x))^6}{6} + C $$

So, the integral of $(\sec(x))^5 \, dx$ is $\frac{(\sec(x))^6}{6} + C$, where $C$ is the constant of integration.