Integral using table and substiution (simple pics provided)

**bikerboy2442** · January 22nd 2013, 09:27 PM

I'm supposed to solve this using an integral table and substitution.

I attempted to solve by rewriting the denominator as Sqrt((8x)²+9²) then using

to solve, substituting u=8x.

Thereby getting the answer ln|8x+sqrt((8x)²+9²)|+C however this is apparently not correct. Any help?

*Edit: My provided integral table can be found here

**Prove It** · January 22nd 2013, 10:18 PM

I would do the substitution $\displaystyle \begin{align*} x = \frac{9}{8}\sinh{t} \implies dx = \frac{9}{8}\cosh{t}\,dt \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{\frac{1}{\sqrt{64x^2 + 81}}\,dx} &= \int{\frac{1}{\sqrt{64\left( \frac{9}{8}\sinh{t} \right)^2 + 81}}\cdot \frac{9}{8}\cosh{t}\,dt} \\ &= \frac{9}{8}\int{\frac{\cosh{t}}{\sqrt{64 \left( \frac{81}{64}\sinh^2{t} \right) + 81 }}\,dt} \\ &= \frac{9}{8}\int{\frac{\cosh{t}}{\sqrt{81\sinh^2{t} + 81}}\,dt} \\ &= \frac{9}{8}\int{\frac{\cosh{t}}{\sqrt{81 \left( \sinh^2{t} + 1 \right) }}\,dt} \\ &= \frac{9}{8}\int{\frac{\cosh{t}}{\sqrt{81\cosh^2{t} }}\,dt} \\ &= \frac{9}{8}\int{\frac{\cosh{t}}{9\cosh{t}}\,dt} \\ &= \frac{1}{8}\int{1\,dt} \\ &= \frac{1}{8}t + C \\ &= \frac{1}{8}\sinh^{-1}{\left( \frac{8x}{9} \right)} + C \end{align*}$

**tom@ballooncalculus** · January 23rd 2013, 03:19 AM

Note that Prove It's answer in log form (Inverse hyperbolic function - Wikipedia, the free encyclopedia) is exactly an eighth of yours after removing the constants.

All you missed was to divide your answer by 8 to adjust for the multiplying by eight caused by the chain rule 'on the way down', i.e. for differentiation.

The conventional way, of course, is to substitute 1/8 du for dx.

But, just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

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