How i do to solve this integrate?

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

2 Answers
Mar 30, 2018

#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#

Mar 30, 2018

#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:
enter image source here
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.