Calc. 2 question! help me please!?

I would rewrite by completing the square:

INT[ sqr(2x-x^2)] dx= INT { sqr[1-(x-1)^2] } dx

Now use trig substitution,

Sin(u)= (x-1)/1= x-1

Sqr[1-(x-1)^2] = cos(u)

X= sinu + 1

Dx = cosu du

INT [ cosu * cosu du]....remember that cos^2(u)= (1/2)(1+ cos(2u))

= INT [(1/2)* (1+ cos2u) ] du

= (1/2) [ u + (1/2)sin2u] + C.....remember that sin2u= 2sinucosu

= (1/2)[ u + sinucosu ] + C

= (1/2) {sin^-1(x-1) + (x-1)* sqr[1-(x-1)^2]} + C

Hoping this helps!

factor out a x.

sqrt(x(2-x)).

INT[sqrtx*sqrt(2-x)]

u^2=x

2udu=dx

INT[u*sqrt(2-u^2)du]

trig sub.

u=sqrt(2)sinv

v=theta

du=sqrt(2)cosv

INT[sqrt(2)sinv*sqrt(2-2sin^2v) *sqrt(2)cosv dv]

INT[2sinvcosv*sqrt(2cos^2v) dv]

INT[2sinvcosv*sqrt(2)*cosv]

2sqrt(2)*INT[2sinvcos^2v dv]

u-sub. I am going to use t.

t=cosv

dt=-sinvdv

-2dt=2sindv

-4sqrt(2)*INT[t^2dt]

-4sqrt(2)*[1/3*t^3]

(-4/3)sqrt(2)*cos^3v.

hower remember

u=sqrt(2)sinv

u/sqrt(2)=sinv

draw a triangle.

cosv=sqrt[(2-u^2)/2)]

however remember

u=sqrtx

so plug that in and you will have your answer.