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

Question

William Abdalla · Accepted Answer

#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#

Emma Aldridge · Answer

undefined To integrate $ \tan^4(x) \sec^4(x) \, dx $, you can use the substitution method. Let $ u = \tan(x) $. Then, $ du = \sec^2(x) \, dx $.

Substitute $ u = \tan(x) $ and $ du = \sec^2(x) \, dx $ into the integral:

$$ \int \tan^4(x) \sec^4(x) \, dx = \int u^4 \, du $$

Integrate $ u^4 $ with respect to $ u $:

$$ \int u^4 \, du = \frac{u^5}{5} + C $$

Now, substitute back $ u = \tan(x) $:

$$ \frac{\tan^5(x)}{5} + C $$

So, the integral of $ \tan^4(x) \sec^4(x) \, dx $ is $ \frac{\tan^5(x)}{5} + C $, where $ C $ is the constant of integration.