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.