Integrating sqrt(10z - z^2) using Trigonometric Substitution

[stunner5000pt]

trying to integrate \sqrt{10z-z^2}

i have been advised to used trig subsitution wherein i subsitute z = cos theta
however i end up at a dead end

\int \sqrt{10 \cos \theta - \cos^2 \theta} sin \theta d \theta

i could certainly replace the sin

\int \sqrt{10 \cos \theta - \cos^2 \theta} \sqrt{1- \cos^2 \theta} d \theta

hwat now do i expand?
or do i use parts?

 

[stunner5000pt]

alternatively i can use z = 10 sin^2 \theta

i am currenttly working o ntaht because i made a mistake... z should be cos theta

is that hter ight way to go??
\int \sqrt{100 sin^2 \theta - 100 \sin^4 \theta} (20 sin \theta cos \theta) dz

 

[GCT]

try z=10-w substitution

 

[arildno]

The simplest choice is to rewrite your root expression as:
\sqrt{10z-z^{2}}=5\sqrt{1-(\frac{z-5}{5})^{2}}

 

[Yegor]

10z-z^2=5^2-(z-5)^2
Try z-5=5\cos{t}

 

[GCT]

what's more simple than this?

I = \int \sqrt{10z-z^2} dz

z=10-t,~dz=-dt

I= - \int (10-t) \sqrt{10-(10-t)} dt

I= -10 \int t^{ \frac{1}{2}} dt + \int t^{ \frac{3}{2}} dt

arildno, can you explain your solution?

 

[Yegor]

I = \int \sqrt{10z-z^2} dz

z=10-t,~dz=-dt

I= - \int (10-t) \sqrt{10-(10-t)} dt

Last line isn't correct. it should be I= - \int \sqrt{t (10-t)} dt

 

[arildno]

It's the same as Yegor's, since we now make the substitution
\frac{z-5}{5}=\cos(t)

 

[GCT]

I see, well with yegor's/arildno proposal, I end up with 25 \int cos^{2} \theta d \theta

 

[Nylex]

I see, well with yegor's/arildno proposal, I end up with 25 \int cos^{2} \theta d \theta

.. then just use \cos 2\theta = 2\cos^{2} \theta - 1, presumably.