How do you integrate #tan^4x sec^4x dx#?

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Answer 1 · 2018-07-23T12:36:49

#inttan^4xsec^4xdx = 1/7tan^7x + 1/5tan^5x + C#

When integrating a function that is a product of tangents and secants, a good strategy is to either to have

#int f(tanx)sec^2x dx#

or to have

#int f(secx)secxtanx dx#.

If we can do this, we can simply integrate by substitution. This may be confusing, but looking at this concrete example will help.

#int tan^4x sec^4x dx#

#= int (tan^4xsec^2x)sec^2x dx#

By the Pythagorean identity #color(red)(sec^2x = tan^2x + 1)#:

#= int (tan^4x(color(red)(tan^2x + 1)))sec^2x dx#

#= int (tan^6x + tan^4x)sec^2x dx#

Now, let #color(blue)(u = tanx)#, #color(blue)(du = sec^2x dx)#.

#= int (color(blue)u^6 + color(blue)u^4)color(blue)(du)#

#= 1/7u^7 + 1/5u^5 + C#

Undoing substitution:

#= 1/7tan^7x + 1/5tan^5x + C#