Is there an error in my textbook?

[Maddie1609]

Homework Statement

Homework Equations

Doesn't the integral of sec x tan x equal sec x?

The Attempt at a Solution

 

[Ray Vickson]

Homework Statement

View attachment 89882

Homework Equations

Doesn't the integral of sec x tan x equal sec x?

The Attempt at a Solution

View attachment 89880

Your textbook is correct. The attachment of your work is too small and blurry for me to read, so I will not even try. I will look at it if you type it out.

 

[Maddie1609]

Your textbook is correct. The attachment of your work is too small and blurry for me to read, so I will not even try. I will look at it if you type it out.

Great, thanks :-) Isn't the integral of sec x tan x equal to sec x?

 

[Mark44]

Great, thanks :-) Isn't the integral of sec x tan x equal to sec x?

Yes, but I don't see how this fact applies to your question.

 

[Ray Vickson]

Great, thanks :-) Isn't the integral of sec x tan x equal to sec x?

Yes, as you can verify by taking the derivative. Likewise, you can check the book's answer by differentiation (which is something you should always do whenever you integrate).

 

[Daeho Ro]

Isn't the integral of sec x tan x equal to sec x?

Well, it is correct, but the textbook uses the fact that \frac{d\tan x}{dx} = \sec^2 x. If you use \frac{d(\sec x \tan x)}{dx} = \sec x, then the first part of integration will be
\int \sec^2 x \tan^2 x dx = \int \sec x \tan^2 x d(\sec x \tan x),
which is not that easy to calculate. However, when you use \frac{d\tan x}{dx} = \sec^2 x, then
\int \sec^2 x \tan^2 x dx = \int \tan^2 x d(\tan x) = \dfrac{1}{3} \tan^3 x + C_1.

 

[Maddie1609]

Well, it is correct, but the textbook uses the fact that \frac{d\tan x}{dx} = \sec^2 x. If you use \frac{d(\sec x \tan x)}{dx} = \sec x, then the first part of integration will be
\int \sec^2 x \tan^2 x dx = \int \sec x \tan^2 x d(\sec x \tan x),
which is not that easy to calculate. However, when you use \frac{d\tan x}{dx} = \sec^2 x, then
\int \sec^2 x \tan^2 x dx = \int \tan^2 x d(\tan x) = \dfrac{1}{3} \tan^3 x + C_1.

Thank you! I just went over it again and realized I forgot about the chain rule.

 

[Maddie1609]

Yes, but I don't see how this fact applies to your question.

I just realized as much myself I forgot about the chain rule, so I intgrated (sec x tan x)² to be (sec³ x)/3.