# Integral.

• Feb 22nd 2010, 05:14 AM
Ted
Integral.
Hello.
evaluate:

$\displaystyle \int \sqrt{x^2+16} dx$.
Without using Trig/Hyperbolic Substitutions.
And you must use the substitution method only.
I believed, its impossible to solve it.
• Feb 22nd 2010, 07:28 AM
The way I was taught to answer integrals:
$\displaystyle \sqrt{x^2+16} dx$

$\displaystyle =(x^2+16)^\frac{1}{2}dx$

$\displaystyle u = x^2+16 => du = 2x$

$\displaystyle =\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]$
$\displaystyle =\frac{1}{2}* \frac{u^{3/2}}{\frac{3}{2}}$

$\displaystyle = \frac{1}{3}(x^2+16)^{3/2}$
• Feb 22nd 2010, 07:35 AM
HallsofIvy
Quote:

The way I was taught to answer integrals:
$\displaystyle \sqrt{x^2+16} dx$

$\displaystyle =(x^2+16)^\frac{1}{2}dx$

$\displaystyle u = x^2+16 => du = 2x$

$\displaystyle =\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]$

Unfortunately, you cannot just put that "x" into the integrand!

Quote:

$\displaystyle =\frac{1}{2}* \frac{u^{3/2}}{\frac{3}{2}}$

$\displaystyle = \frac{1}{3}(x^2+16)^{3/2}$
No, the derivative of $\displaystyle \frac{1}{3}(x^2+ 16)^{3/2}$ is
$\displaystyle (3/2)(1/3)(x^2+ 16)^{1/2} (2x)= x(x^2+16)^{1/2}$, not just $\displaystyle (x^2+ 16)^{1/2}$

I doubt that you were taught that way!

Ted, I don't believe there is any way to do that other than either the trig or hyperbolic substitutions.

Of course, you could always look it up in a table of integrals. Does that count?
http://www.integral-table.com/
• Feb 22nd 2010, 09:06 AM
Krizalid
Quote:

Originally Posted by Ted
Hello.
evaluate:

$\displaystyle \int \sqrt{x^2+16} dx$.
Without using Trig/Hyperbolic Substitutions.

put $\displaystyle x=4u$ and the integral becomes $\displaystyle 16\int\sqrt{u^2+1}\,du.$

put $\displaystyle t=\sqrt{u^2+1}+u$ then $\displaystyle \frac1t=\sqrt{u^2+1}-u,$ thus $\displaystyle \frac12\left(t+\frac1t\right)=\sqrt{u^2+1}$ and $\displaystyle \frac12\left(t-\frac1t\right)=u$ so i think you can finish it.