trigonometric substitution

**Archduke01** · March 25th 2010, 08:17 AM

$\int \frac{\sqrt {4x^2-16}}{x} dx $

$= \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx$

2x = 4 sec t
x = 2 sec t

$\int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx$
$\int \frac {\sqrt {16(sec^2t-1}}{2sect} dx$
$\int \frac {4 \sqrt {sec^2t-1}}{2sect} dx$

How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.

**ione** · March 25th 2010, 08:52 AM

Originally Posted by Archduke01

$\int \frac{\sqrt {4x^2-16}}{x} dx $

$= \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx$

2x = 4 sec t
x = 2 sec t

$\int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx$
$\int \frac {\sqrt {16(sec^2t-1}}{2sect} dx$
$\int \frac {4 \sqrt {sec^2t-1}}{2sect} dx$

How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.

$sec^2t-1=tan^2t$

$dx=2sect tant dt$

**Archduke01** · March 25th 2010, 09:04 AM

Originally Posted by ione

$sec^2t-1=tan^t$

Thanks. So I finally end up with;

$ 2 \int \frac {tan t}{sec t} $

I'm not certain here, but does it become;

$ 2 \frac {-ln |cos t|}{ln |sec t + tan t|} + C ? $

**Grandad** · March 25th 2010, 09:18 AM

Hello Archduke01

Originally Posted by Archduke01

$\int \frac{\sqrt {4x^2-16}}{x} dx $

$= \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx$

2x = 4 sec t
x = 2 sec t

$\int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx$
$\int \frac {\sqrt {16(sec^2t-1}}{2sect} dx$
$\int \frac {4 \sqrt {sec^2t-1}}{2sect} dx$

How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.

You don't need the trig substitution, if you put

$4x^2 - 16 = u^2$

$\Rightarrow x\,dx = \tfrac14u\,du$

Then:

$\int \frac{\sqrt {4x^2-16}}{x} dx$

$=\int \frac{\sqrt {4x^2-16}}{x^2}\,x\,dx$

$=\int\frac{u}{\tfrac14(u^2+16)}\;\tfrac14u\,du$

$=\int\frac{u^2}{u^2+16}\;du$

$=\int\left(1-\frac{16}{u^2+16}\right)\;du$

Can you complete it now?

Grandad

**Archduke01** · March 25th 2010, 09:24 AM

I know trig sub isn't necessary, but I've been asked to use it since it's the new topic. Thank you for your time, though.

**ione** · March 25th 2010, 09:33 AM

Originally Posted by Archduke01

Thanks. So I finally end up with;

$ 2 \int \frac {tan t}{sec t} $

I'm not certain here, but does it become;

$ 2 \frac {-ln |cos t|}{ln |sec t + tan t|} + C ? $

$ 2 \int \frac {tan t}{sec t}dx $

$dx=2sect\,tant\,dt$

**Archduke01** · March 25th 2010, 09:43 AM

Originally Posted by ione

$ 2 \int \frac {tan t}{sec t}dx $

$dx=2sect\,tant\,dt$

Thanks broseidon. I end up with;

$ 2 \int \frac {tan t}{sec t} 2 sect tan t dt $
$= 4 \int tan t tan t dt$
$ = 4 \int tan^2t dt $

I'm sorry for the continuous questions, but how can I integrate that value? I can't find a rule to integrate tan^2 t

**ione** · March 25th 2010, 09:47 AM

Originally Posted by Archduke01

Thanks broseidon. I end up with;

$ 2 \int \frac {tan t}{sec t} 2 sect tan t dt $
$= 4 \int tan t tan t dt$
$ = 4 \int tan^2t dt $

I'm sorry for the continuous questions, but how can I integrate that value? I can't find a rule to integrate tan^2 t

Use $tan^2t=sec^2t-1$

**Archduke01** · March 25th 2010, 10:01 AM

So it becomes;

$4 \int (sec^2t + 1) dt$
$= 4(tan t + t) + C$ ?

**ione** · March 25th 2010, 10:06 AM

Originally Posted by Archduke01

So it becomes;

$4 \int (sec^2t + 1) dt$
$= 4(tan t + t) + C$ ?

Write your answer in terms of x

**Archduke01** · March 25th 2010, 10:20 AM

Right so I have;

4 (tan x + x) + C

Now I've got to construct that triangle. Since originally my x = 2 sec t, it becomes sec t = x/2. If we're looking for the cos it becomes cos t = 2/x. So on the triangle, the hypotenuse is x while the two sides are 2 and $\sqrt {x^2-4}.$

But it's wrong apparently...

**ione** · March 25th 2010, 10:39 AM

Originally Posted by Archduke01

Right so I have;

4 (tan x + x) + C

Now I've got to construct that triangle. Since originally my x = 2 sec t, it becomes sec t = x/2. If we're looking for the cos it becomes cos t = 2/x. So on the triangle, the hypotenuse is x while the two sides are 2 and $\sqrt {x^2-4}.$

But it's wrong apparently...

I find it helpful to draw a right triangle when I begin using trig substitution

Label one of the acute angles t

Since

, label the hypotenuse 2x and the adjacent side 4

Then the opposite side will be

**Archduke01** · March 25th 2010, 10:52 AM

Originally Posted by ione

I find it helpful to draw a right triangle when I begin using trig substitution

Label one of the acute angles t

Since

, label the hypotenuse 2x and the adjacent side 4

Then the opposite side will be

I'm supposed to go to my integrated equation which is 4 (tan x + x) + C.

Using the triangle, I can now replace the tan x with numbers. So my final answer is;

$ 4( \frac {\sqrt{4x^2-16}}{4} + x) + C $

Unfortunately, it's apparently wrong and even after double checking I can't figure out where I messed up.

**ione** · March 25th 2010, 01:35 PM

Originally Posted by Archduke01

So it becomes;

$4 \int (sec^2t + 1) dt$
$= 4(tan t + t) + C$ ?

$4 \int (sec^2t - 1) dt$

$= 4(tan t - t) + C$

$=4tan t - 4t + C$

$=\sqrt{4x^2-16} - 4 arcsec(\frac{x}{2}) + C$

**Grandad** · March 25th 2010, 01:43 PM

Hello Archduke01

Originally Posted by Archduke01

So it becomes;

$4 \int (sec^2t + 1) dt$
$= 4(tan t + t) + C$ ?

You have a sign wrong.

$4\int \tan^2t\;dt$

$=4\int(\sec^2t -1)\;dt$

$=4(\tan t - t) + c$

$= 4(\sqrt{\sec^2t-1}-t) + c$

$= 4\big(\sqrt{\tfrac14x^2-1}-\text{arcsec}(\tfrac12x)\big)+c$

$=2\sqrt{x^2-4}-4\text{ arcsec}(\tfrac12x) + c$

This answer is correct, but there are other ways to write it. So, check it by differentiating, noting that:

$\frac{d}{dx}\big(\text{ arcsec}(\tfrac12x)\big) = \frac{2}{x\sqrt{x^2-4}}$

Grandad

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