Question

Could someone please help me with the integration of this, please?
int_{1}^{4}sqrt{1+(\frac{ 1 }{2sqrt{y} }-9)^2} (see the equation in the comments)

Answer 1

\[\int\limits_{1}^{4}\sqrt{1+(\frac{ 1 }{2\sqrt{y} }-9)^2}\]

Answer 2

I'm wondering if a trig sub might be useful here:
\[\frac{1}{2\sqrt y}-9=\tan u~~~~\iff~~~~\frac{1}{y}=4(\tan u+9)^2\\
-\frac{1}{4y^{3/2}}dy=\sec u \tan u~du\\
-\frac{1}{2y}\left(\frac{1}{2\sqrt y}\right)dy=\sec u \tan u~du\\
-\frac{4(\tan u+9)^2}{2}\left(\tan u+9\right)~dy=\sec u \tan u~du\\
-2(\tan u+9)^3~dy=\sec u \tan u~du\\
dy=-\frac{\sec u \tan u}{2(\tan u+9)^3}~du\]
So the integral would change from
\[{\huge \int}_1^4\sqrt{1+\left(\frac{1}{2\sqrt y}-9\right)^2}~dy\]
to
\[{\huge \int}_?^?\sqrt{1+\tan^2u}~\left(-\frac{\sec u \tan u}{2(\tan u+9)^3}~du\right)\]
Some simplifying:
\[-\frac{1}{2}{\huge \int}_?^?\frac{\sec^2 u \tan u}{(\tan u+9)^3}~du\]
Then another substitution:
\[t=\tan u~~\Rightarrow~~dt=\sec^2 u~du\]
\[-\frac{1}{2}{\huge \int}_?^?\frac{t}{(t+9)^3}~dt\]
I'm not too sure about the limits just yet... and not entirely sure the trig sub is valid, but I'll look it over some.

Answer 3

An integral from hell.
Refer to the Mathematica 9 Home Edition attachment.

Answer 4

thank you both!!