# trigonometric and hyperbolic substitution

• Jun 1st 2008, 11:00 PM
ArTiCK
trigonometric and hyperbolic substitution
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)

ArTiCk
• Jun 2nd 2008, 12:24 AM
angel.white
Quote:

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

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

ArTiCk

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

So unless I really missed something, both answers must be wrong.
• Jun 2nd 2008, 01:08 AM
ArTiCK
Hi,

It says 'using a hyperbolic or trigonometric substitution'

ArTiCk
• Jun 2nd 2008, 01:29 AM
angel.white
Quote:

Originally Posted by ArTiCK
Hi,

It says 'using a hyperbolic or trigonometric substitution'

ArTiCk

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

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

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

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

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

anti-substitute

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

Regardless of what method you use, you should arrive at the same conclusion. Are you sure you copied the problem correctly?
• Jun 2nd 2008, 01:31 AM
ArTiCK
Quote:

Originally Posted by angel.white
$\displaystyle \int\left(1+4x^2\right)~dx$

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

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

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

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

anti-substitute

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

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
• Jun 2nd 2008, 02:21 AM
Isomorphism
Quote:

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:

$\displaystyle \int \sqrt{1+4x^2} \, dx$
• Jun 2nd 2008, 02:26 AM
ArTiCK
Quote:

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:

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

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
• Jun 2nd 2008, 04:32 AM
mr fantastic
Quote:

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

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

Use Osborn's Rule to get the appropriate 'double angle' formula to deal with this .......
• Jun 2nd 2008, 04:49 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
Substitute $\displaystyle x = \frac{1}{2} \sinh t \Rightarrow dx = \frac{1}{2} \cosh t$.

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

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

For checking purposes:

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

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

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