Integration Problem

**ninjaku** · April 4th 2010, 06:46 PM

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

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.

**skeeter** · April 4th 2010, 06:58 PM

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.

$\frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}$

now integrate

**ninjaku** · April 5th 2010, 12:41 AM

Originally Posted by skeeter

$\frac{\tan{x}}{\sec^2{x}} = \tan{x} \cdot \cos^2{x} = \sin{x} \cdot \cos{x}$

now integrate

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

The first way I did using:

$u=sinx$
$du=cosx$

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

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

Which I found to be the answer when I set :

$u=cosx$
$du=-sinx$

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?

**harish21** · April 5th 2010, 12:50 AM

Originally Posted by ninjaku

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

The first way I did using:

$u=sinx$
$du=cosx$

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

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

Which I found to be the answer when I set :

$u=cosx$
$du=-sinx$

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?

$\frac{sin^2x}{2} + c = \frac{1-cos^2x}{2} + c= \frac{1}{2}-\frac{cos^2x}{2} +c = -\frac{cos^2x}{2} +c'$

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

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

**ninjaku** · April 5th 2010, 01:15 AM

That makes a lot of sense, thanks.

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