How do I find the integral of #f(x)=sec^5(x)#?

Question

Ezekiel Alexander · Accepted Answer

#int sec^5xdx = (2sec^3xtanx + 3secxtanx + 3 ln abs(secx+tanx))/8 +C#

Note that in general for any #n >=1#:

#sec^(2n+1) x = sec^(2n-1)xsec^2x#

and:

#sec^2x = d/dx tanx#

so we can integrate by parts:

#int sec^(2n+1)xdx = int sec^(2n-1)d(tanx)#

#int sec^(2n+1)xdx = sec^(2n-1)xtanx - int tanx d(sec^(2n-1)x)#

#int sec^(2n+1)xdx = sec^(2n-1)xtanx - (2n-1)int tan^2xsec(2n-1)x#

Use now the trigonometric identity:

#tan^2x = sec^2x -1#

to get:

#int sec^(2n+1)xdx = sec^(2n-1)xtanx + (2n-1)int (1-sec^2x)sec^(2n-1)x#

and using the linearity of the integral:

#int sec^(2n+1)xdx = sec^(2n-1)xtanx + (2n-1)int sec^(2n-1)xdx- (2n-1) int sec^(2n+1)xdx#

the integral now appears on both sides and we can solve for it:

#2nint sec^(2n+1)xdx = sec^(2n-1)xtanx + (2n-1)int sec^(2n-1)xdx#

#int sec^(2n+1)xdx = (sec^(2n-1)xtanx)/(2n) + (2n-1)/(2n)int sec^(2n-1)xdx#

So, for #n=2#:

#int sec^5xdx = (sec^3xtanx)/4 + 3/4int sec^3xdx#

and for #n=1#

#int sec^3xdx = (secxtanx)/2 + 1/2int secxdx#

Now:

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

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

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

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

Putting the partial results together:

#int sec^5xdx = (2sec^3xtanx + 3secxtanx + 3ln abs(secx+tanx))/8 +C#

Adam Adkins · Answer

undefined To find the integral of f(x) = sec^5(x), you can use the technique of integration by parts. Let's denote u = sec(x) and dv = sec^4(x) dx. Then, differentiate u to get du = sec(x)tan(x) dx and integrate dv to find v = (1/3)sec^3(x) + (2/3)sec(x) dx. Now, apply the integration by parts formula:

∫ u dv = uv - ∫ v du.

Substitute the values of u, v, du, and dv into the formula:

∫ sec^5(x) dx = (1/3)sec^3(x) + (2/3)sec(x)tan(x) - ∫ ((1/3)sec^3(x) + (2/3)sec(x)tan(x))sec(x)tan(x) dx.

Now, simplify the integral:

∫ sec^5(x) dx = (1/3)sec^3(x) + (2/3)sec(x)tan(x) - (1/3)∫ sec^4(x) dx - (2/3)∫ sec^2(x) dx.

The integrals of sec^4(x) and sec^2(x) can be evaluated using trigonometric identities. After evaluating those integrals and simplifying the expression, you will obtain the final result for the integral of f(x) = sec^5(x).