How do you integrate ${tan}^{4} x {sec}^{4} x d x$ $tan^4x sec^4x dx$ ?

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Answer 1 · 2018-07-23T12:36:49

$\int {tan}^{4} x {sec}^{4} x d x = \frac{1}{7} {tan}^{7} x + \frac{1}{5} {tan}^{5} x + C$ $inttan^4xsec^4xdx = 1/7tan^7x + 1/5tan^5x + C$

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

$\int f (tan x) {sec}^{2} x d x$ $int f(tanx)sec^2x dx$

or to have

$\int f (sec x) sec x tan x d x$ $int f(secx)secxtanx dx$ .

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

$\int {tan}^{4} x {sec}^{4} x d x$ $int tan^4x sec^4x dx$

$= \int ({tan}^{4} x {sec}^{2} x) {sec}^{2} x d x$ $= int (tan^4xsec^2x)sec^2x dx$

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

$= \int ({tan}^{4} x ({tan}^{2} x + 1)) {sec}^{2} x d x$ $= int (tan^4x(color(red)(tan^2x + 1)))sec^2x dx$

$= \int ({tan}^{6} x + {tan}^{4} x) {sec}^{2} x d x$ $= int (tan^6x + tan^4x)sec^2x dx$

Now, let $u = tan x$ $color(blue)(u = tanx)$ , $d u = {sec}^{2} x d x$ $color(blue)(du = sec^2x dx)$ .

$= \int (u^{6} + u^{4}) d u$ $= int (color(blue)u^6 + color(blue)u^4)color(blue)(du)$

$= \frac{1}{7} u^{7} + \frac{1}{5} u^{5} + C$ $= 1/7u^7 + 1/5u^5 + C$

Undoing substitution:

$= \frac{1}{7} {tan}^{7} x + \frac{1}{5} {tan}^{5} x + C$ $= 1/7tan^7x + 1/5tan^5x + C$

How do you integrate tan4xsec4xdxtan^4x sec^4x dx?