2 Attachment(s)
Integral using table and substiution (simple pics provided)
Attachment 26671I'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 Attachment 26672 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
Re: Integral using table and substiution (simple pics provided)
I would do the substitution
and the integral becomes
^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*})
Re: Integral using table and substiution (simple pics provided)
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...
http://www.ballooncalculus.org/draw/.../thirtytwo.png
... where (key in spoiler) ...
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