What's the integral of #int tan^5(x) dx#?

2 Answers
Nov 2, 2015

#1/4 sec^4x-sec^2x+ln abs secx#

Explanation:

#int tan^5x dx = int sin^5x/cos^5x dx#

# =int sin^4x(cosx)^-5 sinx dx#

# =int (1-cos^2x)^2(cosx)^-5 sinx dx#

# =int (1-2cos^2x + cos^4x)(cosx)^-5 sinx dx#

# =int ((cosx)^-5 -2(cosx)^-3 +(cosx)^-1) sinx dx#

Now integrate by substitution with #u = cosx#

Alternatively, one could use the reduction formula for #int tan^nx dx#. (If one has it memorized.)

Nov 2, 2015

Another method : )

#inttan^5(x)dx = inttan^2(x)*tan^3(x)dx =int (tan^2(x)+1-1)tan^3(x)dx#

#=int(tan^2+1)tan^3(x)dx-inttan^3(x)dx#

#=int(tan^2+1)tan^3(x)dx-int(tan^2(x)+1)tan(x)dx-inttan(x)dx# (By the same method)

For the two first integral let's #t=tan(x)#

So #dt = (1+tan^2(x))dx#

We have #intt^3dt - inttdt-inttan(x)dx#

#[1/4t^4]-[1/2t^2]+[ln(|cos(x)|)]+C#

(Don't forget absolute value, #ln(f(x))# can't take negative argument !)

Substitute back

#[1/4tan^4(x)]-[1/2tan^2(x)]+[ln(|cos(x)|)]+C#