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:

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

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