OpenStudy (anonymous):

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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thank you both!!