What is the issue with u-Substitution in Integrating tan x sec^4 x dx?

[andrewjohnsc]

Homework Statement

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

Homework Equations

U-Sub

The Attempt at a Solution

I did this two different ways and go two different answers.

The correct way is
$$\int$$sec^3(x)*sec(x)*tan(x) dx
letting u= sec(x) and du= sec(x)*tan(x)
= $$\int$$ u^3 du = sec^4(x)/4 + C
Which is good,

but I can't figure out what's wrong with this way:
$$\int$$tan(x)*sec^4(x) dx
= $$\int$$ tan(x)*(1+tan^2(x))(sec^2(x)) dx
from the identity sec^2(x) = 1 + tan^2(x)
then letting u= tan(x) and du= sec^2(x) dx
gives $$\int$$ u*(1 + u^2) du = $$\int$$ u + u^3 du = u^2/2 + u^4/4 + C
= tan^2(x)/2 + tan^4(x)/4 + C

Which is a different answer from the correct one. What's wrong with the second way? Does it have to do with the u-sub... Thanks,
Andrew

 

[Cyosis]

Both answers are correct, but the constant C in your second answer isn't the same as the constant C in your first answer.

$$\tan^2(x)/2 + \tan^4(x)/4+C'=\frac{1}{4}(\sec^4(x)-1)+C'=\sec^4(x)/4-1/4+C'=\sec^4(x)+A$$

I leave the intermediate steps to you.

 

[rock.freak667]

sec⁴x=(1+tan²)²

$$sec^4x= 1 + 2tan^2x+tan^4x$$

Now divide by 1/4 and you'll see that they are the exact same thing.

AND 1/4+C = another constant!

 

[andrewjohnsc]

Thanks a lot!