How to [itex]\int{\frac{-2x+20}{2\sqrt{-x^2+20x}}}dx[/itex]

[Ocata]

Hello,

What would be a good method, for a novice in integration, to solve this integral. I haven't done much integration, so this looks quite daunting. Would this be solvable with trig substitution by itself or would I need to employ partial fractions?

\int{\frac{-2x+20}{2\sqrt{-x^2+20x}}}dx

If I complete the square in the denominator, I wonder if I could make it easier.

\int{\frac{-2x+20}{2\sqrt{-(x-10)^2-10}}}dx

What would be a logical method?

 

[phion]

Try a u-substitution.

 

[phion]

\frac{1}{\sqrt{u}}

 

[Ocata]

Hi,

I found a similar example in the Stewart calculus book, chapter 7.3 example 7. I completed the square in the denominator, did u-substitution u=x-10, then trig substitution u= 10sin(theta).

The integratable term I arrived at was \int (-10sin(theta)) = 10cos(theta)

Then I substituted back in the x terms to get y(x)=\sqrt{10^2 -(x-10)^2)} => ((x-10)^2)+y^2=10^2

Which is a circle with radius 10 and I believe with center at the x-axis and left side passing through the origin.

I think this is correct as it is the equation I initially started out with and took the derivative of before integrating back.

(note: I apologize, I don't have the latex avalable on my phone. I did my best and will fix any errors when I get to a computer at school in a few hours.)

 

[phion]

Hi,

I found a similar example in the Stewart calculus book, chapter 7.3 example 7. I completed the square in the denominator, did u-substitution u=x-10, then trig substitution u= 10sin(theta).

The integratable term I arrived at was \int (-10sin(theta)) = 10cos(theta)

Then I substituted back in the x terms to get y(x)=\sqrt{10^2 -(x-10)^2)} => ((x-10)^2)+y^2=10^2

Which is a circle with radius 10 and I believe with center at the x-axis and left side passing through the origin.

I think this is correct as it is the equation I initially started out with and took the derivative of before integrating back.

(note: I apologize, I don't have the latex avalable on my phone. I did my best and will fix any errors when I get to a computer at school in a few hours.)

You over-complicated things a little bit, but nice job!

 

[Byeonggon Lee]

You can also use u-substitution as phion said$\int\frac{-2x+20}{\sqrt{-x^2+20}}dx$

As you can see

-2x+20 is derivative of -x^2+20

And if you substitute -x^2+20 with variable such as t, then the problem becomes simpler

$t=-x^2+20x$

$\frac{d(t)}{dt} = \frac{d(-x^2+20x)}{dt}$

$1 = (-2x+20) \cdot \frac{dx}{dt}$

$dt = (-2x+20) \cdot dx$

You can substitue $(-2x+20) \cdot dx$ with $dt$
and $-x^2+20$ with $t$

then

$\int\frac{-2x+20}{\sqrt{-x^2+20}}dx$

becomes

$\int\frac{1}{2\sqrt{t}} dt$

$=\frac{1}{2}\int t^{-\frac{1}{2}} dt$

$=t^{\frac{1}{2}}$

 

[Ocata]

Thank you.

Wow, that was a much more efficient choice of substitution. There must be many choices for substituting in the integrand. Most of which are overly complicated and a few of which are more elegant. Thanks for shedding light on this topic.

Regards