Question

How do you find the integral of `sqrt (2x - x^2) dx`?

Answer 1

The answer is `=1/2arcsin(x-1)+1/2(x-1)sqrt(2x-x^2)+C`

Explanation:

`2x-x^2=1-(x-1)^2` by completing the square.

Therefore, the integral is

`I=intsqrt(2x-x^2)dx=intsqrt(1-(x-1)^2)dx`

Let `u=x-1`, `=>`, `du=dx`

`I=intsqrt(1-u^2)du`

Let `u=sintheta`, `=>`, `du=costhetad theta`

`sqrt(1-u^2)=sqrt(1-sin^2theta)=costheta`

`I=intcostheta*costhetad theta=intcos^2theta d theta`

`cos2theta=2cos^2theta-1`

`=>`, `cos^2theta=(1+cos2theta)/2`

Therefore,

`I=1/2int(1+cos2theta)d theta`

`=1/2(theta+1/2sin2theta)`

`=1/2theta+1/4sin2theta`

`=1/2theta+1/2sinthetacostheta`

`=1/2arcsin(u)+1/2usqrt(1-u^2)`

`=1/2arcsin(x-1)+1/2(x-1)sqrt(1-(x-1)^2)+C`

`=1/2arcsin(x-1)+1/2(x-1)sqrt(2x-x^2)+C`

Answer 2

`arcsinsqrt(x/2)-1/2(1-x)sqrt(2x-x^2)+C`.

Explanation:

Here is Second Method :

Let, `x=2sin^2u. :. dx=2*2sinu*cosudu`.

`:. intsqrt(2x-x^2)dx`,

`=intsqrt(4sin^2u-4sin^4u)(4sinucosu)du`,

`=4intsqrt{4sin^2u(1-sin^2u)}sinucosudu`,

`=4int(2sinucosu)sinucosudu`,

`=2int(4sin^2ucos^2u)du`,

`=2intsin^2 2udu`,

`=2int(1-cos4u)/2du`,

`=u-1/4sin4u`,

`=u-1/4*2sin2ucos2u`,

`=u-1/2(2sinucosu)(1-2sin^2u)`,

`=u-sqrt{sin^2ucos^2u}(1-2sin^2u)`,

`=u-sqrt{sin^2u(1-sin^2u)}(1-2sin^2u)`.

Since, `x=2sin^2u, sinu=sqrt(x/2). :. u=arcsinsqrt(x/2)`.

`rArr intsqrt(2x-x^2)dx`,

`=arcsinsqrt(x/2)-sqrt{x/2(1-x/2)}(1-x)`,

`=arcsinsqrt(x/2)-1/2(1-x)sqrt(2x-x^2)+C`.

Enjoy Maths.!