Question

Question #2c6b4

Answer 1

`int(sqrt(-x^2+2x+3))dx=4sin^-1((x-1)/2)`

Explanation:

`int(sqrt(-x^2+2x+3))dx`

Complete the square inside the square root
`int(sqrt(-(x^2-2x-3)))dx`

`int(sqrt(-(x-1)^2+4))dx`

Rewritten:
`int(sqrt(4-(x-1)^2))dx`

This format of square root of `sqrt(a^2-x^2)` can be associated with using a [trigonometric substitution](http://calculus.nipissingu.ca/tutorials/integrals.html`trs) for`asintheta#

`sintheta=frac{x-1}{2}`
`2sintheta=x-1`
`x=2sintheta+1`
`theta=sin^-1(frac{x-1}{2})`

`dx/(d theta)=2cos theta`
`dx=2costheta d theta`

`sectheta=sqrt(4-(x-1)^2)/2`
`sqrt(4-(x-1)^2)=2sectheta`

Now substitute these values into the the integral:
`int((2sectheta)2costheta )d theta`

`=int(4)d theta`

`=4theta`

`=4sin^-1(frac{x-1}{2})`

Answer 2

`int sqrt (3+2x-x^2) dx = 2arcsin((x-1)/2) + (x-1)/2sqrt(3+2x-x^2)+C`

Explanation:

We have:

`int sqrt (3+2x-x^2) dx`

Complete the square under the root:

`3+2x-x^2 = 4 - (1-x)^2 = 4 (1- ((1-x)/2)^2)`

and substitute:

`t= (1-x)/2`

`dt = -dx/2`

so that:

`int sqrt (3+2x-x^2) dx = int sqrt (4 (1- ((1-x)/2)^2))dx = 2int sqrt (1- ((1-x)/2)^2)dx = -4int sqrt(1-t^2)dt`

Substitute again:

`t = sinu`

`dt = cosu du`

and note that as the integrand is defined only for `t in [-1,1]` we have that `u in [-pi/2,pi/2]`:

`int sqrt(1-t^2)dt = int sqrt (1-sin^2u) cosu du = int sqrt (cos^2u) cosu du`

For `u in [-pi/2,pi/2]`, `cos u > 0`, then:

`int sqrt(1-t^2)dt = int cos^2udu`

Use now the trigonometric identity:

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

`int cos^2udu = int (1+cos2u)/2 du = 1/2 int du + 1/4 int cos(2u)d(2u) = u/2 + sin (2u)/4+C = 1/2(u+sinucosu)+C`

Undo now the substitution:

`u = arcsint`

`sin u = t`

`cos u = sqrt(1-t^2)`

so:

`int sqrt(1-t^2)dt = 1/2(arcsint +tsqrt(1-t^2))+C`

and:

`int sqrt (3+2x-x^2) dx = -2(arcsin((1-x)/2) + ((1-x)/2)sqrt(1-((1-x)/2)^2))+C`

simplifying:

`int sqrt (3+2x-x^2) dx = 2arcsin((x-1)/2) + (x-1)/2sqrt(3+2x-x^2)+C`