Question

How i do to solve this integrate?

`int(t-2)/sqrt(27+6t-t^2)dt`

Answer 1

`int(t-2)/(sqrt(27+6t-t^2))dt`

`=-sqrt(36-(t-3)^2)+arcsin((t-3)/6)+C`

Explanation:

We are asked to evaluate the indefinite integral

`int(t-2)/(sqrt(27+6t-t^2))dt`

Let's begin by completing the square under the radical.

`rArrint(t-2)/(sqrt(27-(t^2-6t)))dt`

`rArrint(t-2)/(sqrt(27-(t^2-6t+color(red)(9)-color(orange)9)))dt`

`rArrint(t-2)/(sqrt(27+color(orange)9-(t^2-6t+color(red)(9))))dt`

`rArrint(t-2)/(sqrt(36-(t-3)^2))dt`

We can now apply trigonometric substitution.

Let

`sintheta=(t-3)/6`

`rArrt-3=6sintheta`

`rArrcolor(red)(t-2=6sintheta+1)`

`rArrcolor(green)(dt=6costheta)` `color(green)d` `color(green)theta`

And let

`costheta=(sqrt(36-(t-3)^2))/6`

`rArrcolor(orange)(sqrt(36-(t-3)^2)=6costheta)`

Now we can rewrite the integral in terms of `theta`

`int` `(color(red)(6sintheta+1))/(color(orange)(6costheta))color(green)(6costheta)` `color(green)d` `color(green)theta`

This greatly simplifies the integral!

`int` `(color(red)(6sintheta+1))/(cancelcolor(orange)(6costheta))cancelcolor(green)(6costheta)` `color(green)d` `color(green)theta`

`rArrint` `(6sintheta+1)` `d` `theta`

Integrating, we get

`rArr-6costheta+theta+C`

And now we can rewrite the result in terms of `t`

`rArr-6((sqrt(36-(t-3)^2))/6)+arcsin((t-3)/6)+C`

`rArr-sqrt(36-(t-3)^2)+arcsin((t-3)/6)+C`

Answer 2

`int(t-2)/sqrt(27+6t-t^2)dt=2arcsin(sqrt(3(t+3))/6)-sqrt(27+6t-t^2)+C`

Explanation:

.

`int(t-2)/sqrt(27+6t-t^2)dt=int(t-2)/sqrt((-(t^2-6t-27)))dt=`

`int(t-2)/sqrt(-(t-9)(t+3))dt`

Let `u=t+3, :. du=dt`, and

`t=u-3, :. t-2=u-5, t-9=u-12,`

Let's substitute:

`=int(u-5)/sqrt(-(u-12)u)du=int(u-5)/sqrt(u(12-u))du=`

`int(u-5)/sqrt(12u-u^2)du`

Let's use trigonometric substitution:

We draw a right triangle and label an angle `theta` as shown below:

Now, we will right the basic trigonometric functions for the angle `theta`:

`sintheta=u/sqrt(12u)=sqrt(12u)/12`

`costheta=sqrt(12u-u^2)/sqrt(12u)`

`tantheta=u/sqrt(12u-u^2`

`sin^2theta=u^2/(12u)=u/12`

`2sinthetacosthetad theta=(du)/12`

`u=12sin^2theta`

`du=24sinthetacosthetad theta`

`sqrt(12u-u^2)=sqrt(12u)costheta=sqrt(12(12sin^2theta))costheta`

`sqrt(12u-u^2)=12sinthetacostheta`

Let's substitute:

`int(u-5)/sqrt(12u-u^2)du=int(12sin^2theta-5)/(12sinthetacostheta)(24sinthetacostheta)d theta=`

`2int(12sin^2theta-5)d theta=24intsin^2thetad theta-10intd theta=`

`24int(1-cos2theta)/2d theta-10theta=12intd theta-12intcos2thetad theta-10theta=`

`12theta-12I-10theta=2theta-12I`

`I=intcos2thetad theta`

Let `z=2theta, :. dz=2d theta, :. d theta=(dz)/2`

`I=1/2intcoszdz=1/2sinz=1/2sin2theta`

`2theta-12I=2theta-6sin2theta`

Using the double angle formula:

`sin2x=2sinxcosx`

`2theta-12I=2theta-12sinthetacostheta`

Let's substitute back for `u`:

`int(u-5)/sqrt(12u-u^2)du=2arcsin(sqrt(12u)/12)-12(sqrt(12u)/12)(sqrt(12u-u^2)/sqrt(12u))=`

`2arcsin(sqrt(12u)/12)-cancelcolor(red)(12)(cancelcolor(purple)(sqrt(12u)))/cancelcolor(red)(12)(sqrt(12u-u^2)/cancelcolor(purple)(sqrt(12u)))=`

`=2arcsin((sqrt(3u))/6)-sqrt(12u-u^2`

Now, we can substitute back for `t`:

`int(t-2)/sqrt(27+6t-t^2)dt=2arcsin(sqrt(3(t+3))/6)-sqrt(27+6t-t^2)+C`

This is one form of the solution. The result can be in many different forms depending on what you choose to be `u` and how you set up your triangle and what trigonometric functions you use.