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)
\[\int\limits_{1}^{4}\sqrt{1+(\frac{ 1 }{2\sqrt{y} }-9)^2}\]
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.
An integral from hell. Refer to the Mathematica 9 Home Edition attachment.
thank you both!!
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