- #1
tshaw101
- 3
- 0
I am having an issue with this problem.
[tex]\int tan(x)^3: [/tex]
You separate a tan(x)^2, and use the identity 1+tan(x)^2 = sec(x)^2
You then end up with [tex]\int tan(x) sec(x)^2 dx[/tex] + [tex]\int tan(x) dx[/tex]
[tex]\int tan(x) dx = ln absval(sec(x)) [/tex]
[tex]\int tan(x) sec(x)^2 dx[/tex]
Here, the book says to solve the problem this way:
set u = tan(x) du = sec(x)^2dx
so [tex]\int u du = 1/2 u ^2 = 1/2 tan(x)^2 [/tex]
Why can't you solve this problem this way instead?
[tex]\int (tan(x) sec(x)) sec(x) dx[/tex]
set u = sec(x), du = (tan(x) sec(x)) dx
so [tex]\int u du = 1/2 u ^2 = 1/2 sec(x)^2 [/tex]
I think I must be missing something...any help you could give me would be greatly appreciated.
[tex]\int tan(x)^3: [/tex]
You separate a tan(x)^2, and use the identity 1+tan(x)^2 = sec(x)^2
You then end up with [tex]\int tan(x) sec(x)^2 dx[/tex] + [tex]\int tan(x) dx[/tex]
[tex]\int tan(x) dx = ln absval(sec(x)) [/tex]
[tex]\int tan(x) sec(x)^2 dx[/tex]
Here, the book says to solve the problem this way:
set u = tan(x) du = sec(x)^2dx
so [tex]\int u du = 1/2 u ^2 = 1/2 tan(x)^2 [/tex]
Why can't you solve this problem this way instead?
[tex]\int (tan(x) sec(x)) sec(x) dx[/tex]
set u = sec(x), du = (tan(x) sec(x)) dx
so [tex]\int u du = 1/2 u ^2 = 1/2 sec(x)^2 [/tex]
I think I must be missing something...any help you could give me would be greatly appreciated.