Solve the Integral: \int\frac{9x}{\sqrt{6x-x^2}}dx

[G01]

This one has me stumped.

\int\frac{9x}{\sqrt{6x-x^2}}dx

I've tried completeing the square in the root and using trig substitution, but that made it really complicated. Also tried to rationalize the denominator, but to know avail. Any hints?

 

[topsquark]

This one has me stumped.

\int\frac{9x}{\sqrt{6x-x^2}}dx

I've tried completeing the square in the root and using trig substitution, but that made it really complicated. Also tried to rationalize the denominator, but to know avail. Any hints?

There might be a fancier way, but try this:
\int\frac{9x}{\sqrt{6x-x^2}}dx

=\int \frac{9x}{\sqrt x \sqrt{6-x}}dx

= \int \frac{9 \sqrt x}{\sqrt{6-x}} dx

Then use a substitution: y = \sqrt{6-x}. You should get a form that you can do a trig substitution on.

-Dan

 

[pizzasky]

We want to change the numerator into something "nicer", so the integration will be easier. Try this...

\frac{9x}{\sqrt{6x-x^2}} = (\frac{-9}{2}) (\frac{(6-2x)-6}{\sqrt{6x-x^2}})

 

[G01]

ok i got this does this seem right?

54\sin^{-1}(\frac{\sqrt{6-x}}{\sqrt{6}}) + \frac{\sqrt{x^2 - 6x}}{6} + C

 

[G01]

wait I am trying your way now pizzaky

 

[G01]

got it your way pizzaky, that was very easy. I am pretty sure its right.

 

[pizzasky]

So basically, if the expression you need to integrate is of the form \frac{linear}{quadratic} or \frac{linear}{\sqrt{quadratic}}, we try to get something like the differentiated version of the quadratic expression into the numerator.

Remember to add a factor in front of the numerator (in this case \frac{-9}{2}) so that your numerator will consist of the differentiated version of the denominator and only a constant, which makes integration easier.

By the way, is your final answer the same as what you wrote above?

 

[dextercioby]

Completing the square and using the substitution x=3\sin t +3 is a very elegant way to do it...

Daniel.