trigonometric and hyperbolic substitution

**ArTiCK** · June 1st 2008, 11:00 PM

Hi all,

My apologises to angel.white, i completely typed the incorrect question and i would like to thank Isomorphism

Using an appropriate trigonometric or hyperbolic substitution, find the indefinite integrals of
sqrt( 1 + 4x^2)

Thanks in advance,
ArTiCk

**angel.white** · June 2nd 2008, 12:24 AM

Originally Posted by ArTiCK

Hi all,

I have a feeling the answer to the following question is incorrect:

Using an appropriate trigonometric or hyperbolic substitution, find the indefinite integrals of
$<br /> \int{{1 + 4x^2}} dx<br />$

I got 1/4 (x sqrt( 1 + 4x^2) + 1/2arcsinh2x) + c...

however the answer says: 1/4(arcsinh(2x) + 2x*sqrt(1 + 4x^2)) + C

Thanks in advance,
ArTiCk

Well, if you don't use a trig/hyperbolic substitution, you get $\int\left(1+4x^2\right)~dx = x+\frac{4x^3}3+C$

So unless I really missed something, both answers must be wrong.

**ArTiCK** · June 2nd 2008, 01:08 AM

Hi,

It says 'using a hyperbolic or trigonometric substitution'

ArTiCk

**angel.white** · June 2nd 2008, 01:29 AM

Originally Posted by ArTiCK

Hi,

It says 'using a hyperbolic or trigonometric substitution'

ArTiCk

$\int\left(1+4x^2\right)~dx$

-------------------------
substitution:
$sin(a) = x$

$cos(a) ~da = ~dx$
-------------------------

$=\int cos(a)\left(1+4sin^2(a)\right)~da$

$=\int cos(a) ~da + 4 \int cos(a)~sin^2(a)~da$

$=sin(a) + \frac 43sin^3(a)+C$

anti-substitute

$=x +\frac 43x^3+C$

Regardless of what method you use, you should arrive at the same conclusion. Are you sure you copied the problem correctly?

**ArTiCK** · June 2nd 2008, 01:31 AM

Originally Posted by angel.white

$\int\left(1+4x^2\right)~dx$

-------------------------
substitution:
$sin(a) = x$

$cos(a) ~da = ~dx$
-------------------------

$=\int cos(a)\left(1+4sin^2(a)\right)~da$

$=\int cos(a) ~da + 4 \int cos(a)~sin^2(a)~da$

$=sin(a) + \frac 43sin^3(a)+C$

anti-substitute

$=x +\frac 43x^3+C$

Regardless of what method you use, you should arrive at the same conclusion. Are you sure you copied the problem correctly?

Yes i am totally sure that i copied it correctly, because i have already done some and have the correct answer.... besides this one

Thanks for trying though
ArTiCk

**Isomorphism** · June 2nd 2008, 02:21 AM

Originally Posted by ArTiCK

Hi,

It says 'using a hyperbolic or trigonometric substitution'

ArTiCk

If it said so, I am nearly convinced that you must have missed a square root sign. Check it again carefully. Polynomial integration does not need 'a hyperbolic or trigonometric substitution'.

Most likely, the question was:

$\int \sqrt{1+4x^2} \, dx$

**ArTiCK** · June 2nd 2008, 02:26 AM

Originally Posted by Isomorphism

If it said so, I am nearly convinced that you must have missed a square root sign. Check it again carefully. Polynomial integration does not need 'a hyperbolic or trigonometric substitution'.

Most likely, the question was:

$\int \sqrt{1+4x^2} \, dx$

I apologize to angel.white and thank you isomorphism for picking up the mistake. The correct equation is in the first post.

Thank you once again,
ArTiCk

**mr fantastic** · June 2nd 2008, 04:32 AM

Originally Posted by ArTiCK

I apologize to angel.white and thank you isomorphism for picking up the mistake. The correct equation is in the first post.

Thank you once again,
ArTiCk

Substitute $x = \frac{1}{2} \sinh t \Rightarrow dx = \frac{1}{2} \cosh t\, dt$ .

Then the integral becomes $\frac{1}{2} \int \cosh^2 t \, dt .....$

Use Osborn's Rule to get the appropriate 'double angle' formula to deal with this .......

**mr fantastic** · June 2nd 2008, 04:49 AM

Originally Posted by mr fantastic

Substitute $x = \frac{1}{2} \sinh t \Rightarrow dx = \frac{1}{2} \cosh t$ .

Then the integral becomes $\frac{1}{2} \int \cosh^2 t \, dt .....$

Use Osborn's Rule to get the appropriate 'double angle' formula to deal with this .......

For checking purposes:

$x = \frac{1}{2} \sinh t \Rightarrow t = \text{arcsinh} \, (2x)$ . Therefore:

$\frac{1}{4} t \rightarrow \frac{1}{4} \text{arcsinh} \, (2x)$ .

$\frac{1}{8} \sinh (2t) = \frac{1}{4} \sinh t \cosh t \rightarrow \frac{1}{4} (2x) \sqrt{1 + 4x^2} = \frac{1}{2} \, x \, \sqrt{1 + 4x^2}$ .

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