+0

# integration

0
1418
11
+1832

Feb 3, 2015

#2
+26319
+15

$$\int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\int \frac{1}{\sqrt{4[ 1- \left( \frac{x-3}{2} \right)^2} ] }\ dx =\frac{1}{2}\int \frac{1}{\sqrt{1- \left( \frac{x-3}{2} \right)^2}}\ dx\\\\ \small{\text{We substitute:  \frac{x-3}{2}=\sin{(u)},  so we have  \frac{1}{2}\ dx=\cos{(u)}\ du  or  \ dx=2\cos{(u)}\ du  }}\\ =\frac{1}{2}\int \frac{2\cos{(u)}}{\sqrt{1- \sin^2{(u)}}}\ du =\frac{1}{2}\int \frac{2\cos{(u)}}{\cos{(u)}}}\ du =\int\ du=u\\ \small{\text{We substitute back:  u=\arcsin{ ( \frac{x-3}{2} ) }  }}\\\\ \int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\sin^{-1}{ ( \frac{x-3}{2} ) }$$

Feb 3, 2015

#1
+124524
+15

-5 + 6x - x^2 =

-5 - (x^2 - 6x )          complete the square on x...take one-half the coefficient on x

(= 3), square it (= 9)....then......add and subtract it...so we have...

-5 - (x^2 - 6x + 9 - 9) =

-5 + 9 - (x^2 - 6x + 9)       factor  the expression on the parenthesis

4 - (x - 3)^2

And that's it...

Feb 3, 2015
#2
+26319
+15

$$\int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\int \frac{1}{\sqrt{4[ 1- \left( \frac{x-3}{2} \right)^2} ] }\ dx =\frac{1}{2}\int \frac{1}{\sqrt{1- \left( \frac{x-3}{2} \right)^2}}\ dx\\\\ \small{\text{We substitute:  \frac{x-3}{2}=\sin{(u)},  so we have  \frac{1}{2}\ dx=\cos{(u)}\ du  or  \ dx=2\cos{(u)}\ du  }}\\ =\frac{1}{2}\int \frac{2\cos{(u)}}{\sqrt{1- \sin^2{(u)}}}\ du =\frac{1}{2}\int \frac{2\cos{(u)}}{\cos{(u)}}}\ du =\int\ du=u\\ \small{\text{We substitute back:  u=\arcsin{ ( \frac{x-3}{2} ) }  }}\\\\ \int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\sin^{-1}{ ( \frac{x-3}{2} ) }$$

heureka Feb 3, 2015
#3
+1832
+5

But where is my mistake here

Feb 5, 2015
#4
+124524
+10

Let's go from this step

-(x^2 -6x + 9 - 9) - 5    =

-(x^2 - 6x + 9) +  9 - 5  =

-(x - 3)^2 + 4  =

4 - (x - 3)^2

Feb 5, 2015
#5
+1832
+5

But why we cant multiply the negative sign inside the parentheses in this step

-(x^2 - 6x + 9) +  9 - 5

so its become

(-x^2 + 6x - 9) + 9 - 5

Feb 6, 2015
#6
+117762
+10

You mistake is that

$$(x-3)^2=x^2-6x+9\\\\ It does not equal -x^2+6x-9\\\ that is why -1 had to be left out of the bracket!$$\$

Feb 6, 2015
#7
+124524
+5

Thanks, Melody.....

Feb 6, 2015
#8
+117762
+5

My pleasure Chris :)

Feb 6, 2015
#9
+1832
0

Melody

I don't get it

$$-x^2+6x-9$$ = $$(x-3)^2$$

Also

$$x^2-6x+9 = (x-3)^2$$

so what is wrong !!!

Feb 18, 2015
#10
+117762
+5

NO

$$-x^2+6x-9$$  does not equal      $$(x-3)^2$$

$$\\-x^2+6x-9= -(x^2-6x+9)=-1*(x^2-6x+9) = -1*(x-3)^2\\\\ the -1 is not squared so \\\\ (x-3)^2 \ne -x^2+6x-9$$

remember

$$-3^2\ne(-3)^2$$

consider these two sums

$$20+-3^2 = 20-3^2=20-9=11$$

and

$$20+ (-3)^2 = 20++9=29$$

see        $$-3^2\ne(-3)^2$$

I hope that helps :)

Feb 18, 2015
#11
+1832
+5

Now its clear .. thank you

Feb 18, 2015