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.

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- Feb 22nd 2010, 05:14 AMTedIntegral.
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 AMlil_cookie
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}}

$

Answer:

$\displaystyle

= \frac{1}{3}(x^2+16)^{3/2}

$ - Feb 22nd 2010, 07:35 AMHallsofIvy
Unfortunately, you cannot just put that "x" into the integrand!

Quote:

$\displaystyle

=\frac{1}{2}*

\frac{u^{3/2}}{\frac{3}{2}}

$

Answer:

$\displaystyle

= \frac{1}{3}(x^2+16)^{3/2}

$

$\displaystyle (3/2)(1/3)(x^2+ 16)^{1/2} (2x)= x(x^2+16)^{1/2}$, 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 AMKrizalid
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.