Integrating Trig Functions: How to Solve for the Integral of tan x sec2x?

[hasan_researc]

Homework Statement

What is the integral of tan x sec²x with respect to x?

Homework Equations

The Attempt at a Solution

I have no idea as to how I should proceed!

 

[hikaru1221]

Hint: [tex]tanxsec^2xdx=\frac{sinxdx}{cos^3x}[/tex]

 

[hasan_researc]

How can that help? I have no idea!

Also, if we let u = tan x, then we get the limit of sin x as x tends to infinity, which is nonsense.

 

[hikaru1221]

Another hint: sinxdx = d( ... )?

 

[HallsofIvy]

How can that help? I have no idea!

Also, if we let u = tan x, then we get the limit of sin x as x tends to infinity, which is nonsense.

As you were told in another thread, the "indefinite" integral is just the anti-derivative. It has nothing to do with a limit at infinity.
To integrate
[tex]\int \frac{sin x}{cos^3 x} dx[/tex]
Let u= cos(x).

 

[Mark44]

Even more direct: If u = tanx, du = sec²x dx. The indefinite integral has the form [itex]\int u du[/itex].

 

[hasan_researc]

I don't understand what d(...) actually means. I guess it's a clever way of using calculus that I'm not familiar with. But I have used the substitution u = cos x as follows.

[tex]
u = \cos x & \Rightarrow du = - sin x dx \\
\int \frac{\sin x dx}{cos^{3} x} & = - \int\frac{1}{u^3} du \\
& = \frac{1}{2} u^{-2} + c \\
& = \frac{1}{2\cos^{2}x} + c
[/tex]

But if I use u = tanx I get the following.

[tex]
u = \tan x & \Rightarrow du = sec^{2} x dx \\
\int tan x sec^{2} x dx & = - \intu du \\
& = \frac{1}{2} u^{2} + c \\
& = \frac{tan^{2} x}{2} + c

[/tex]

The two answers are contradictory. Where's the problem?

 

[hasan_researc]

Sorry I made a silly mistake in my Latex code. The correction is:

[tex]
\int tan x sec^{2} x dx & = \int u du \\
[/tex]

 

[hasan_researc]

And I don't know how to break lines in my Latex code. Sorry for that!

 

[vela]

The two answers are contradictory. Where's the problem?

They're not contradictory. Use the identity tan² x + 1 = sec² x.

 

[hasan_researc]

Ok, so

[tex]
\frac{1}{2}\sec^{2} x + c \\
& = \frac{1}{2}\tan^{2} x + \frac{1}{2} + c \\
[/tex]

Therefore, the constant of integration resulting from my math is 1/2 + c, whereas the constant of integration in the other result is c. Should we not be worried abt that? Or is it simply an effect of the use of different substitutions at the start of the problem?

 

[Mark44]

If you get two different answers from an indefinite integral, they can differ by only a constant. (1/2)sec^2(x) and (1/2)tan^2(x) differ by a constant, which is what vela was saying.