# Thread: trigonometric and hyperbolic substitution

1. ## 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

2. 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
$
\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 $\int\left(1+4x^2\right)~dx = x+\frac{4x^3}3+C$

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

3. Hi,

It says 'using a hyperbolic or trigonometric substitution'

ArTiCk

4. 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?

5. 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

6. 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:

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

7. 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

8. 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 .......

9. 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:

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

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