1. Integration Problem

Find the integral of $\displaystyle (tanx/sec^2x)dx$

I tried using u substitution, and I get:

$\displaystyle u=tanx$
$\displaystyle du=(sec^2)x$

At first it seems like it was going to be simple.

But in the problem there is a $\displaystyle sec^-2$
And in my substitution there is a $\displaystyle sec^2$.

Any help would be appreciated.

2. Originally Posted by ninjaku
Find the integral of (tanx/(sec^2)x)dx.

I tried using u substitution, and I get:

u=tanx
du=(sec^2)x.

At first it seems like it was going to be simple.

But in the problem there is a sec^-2.
And in my substitution there is a sec^2.

Any help would be appreciated.
$\displaystyle \frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}$

now integrate

3. Originally Posted by skeeter
$\displaystyle \frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}$

now integrate
Ok after your suggestion. I did the problem two ways.

The first way I did using:

$\displaystyle u=sinx$
$\displaystyle du=cosx$

and got the answer to be $\displaystyle \frac{sin^2x}{2} + c.$

However when I checked it in my calculator, it said the answer was instead, $\displaystyle -\frac{cos^2x}{2} +c.$

Which I found to be the answer when I set :

$\displaystyle u=cosx$
$\displaystyle du=-sinx$

I would just like to know if both of those are correct, or only one.
And if only one of them is correct, how would I know which term I should pick to use for the substitution?

4. Originally Posted by ninjaku
Ok after your suggestion. I did the problem two ways.

The first way I did using:

$\displaystyle u=sinx$
$\displaystyle du=cosx$

and got the answer to be $\displaystyle \frac{sin^2x}{2} + c.$

However when I checked it in my calculator, it said the answer was instead, $\displaystyle -\frac{cos^2x}{2} +c.$

Which I found to be the answer when I set :

$\displaystyle u=cosx$
$\displaystyle du=-sinx$

I would just like to know if both of those are correct, or only one.
And if only one of them is correct, how would I know which term I should pick to use for the substitution?
$\displaystyle \frac{sin^2x}{2} + c = \frac{1-cos^2x}{2} + c= \frac{1}{2}-\frac{cos^2x}{2} +c = -\frac{cos^2x}{2} +c'$

Note: $\displaystyle c' = c + \frac{1}{2}$

these two terms $\displaystyle \frac{sin^2x}{2} + c$ and $\displaystyle -\frac{cos^2x}{2} +c'$ differ only by a constant. Either answer works in this case.

5. That makes a lot of sense, thanks.