How do you integrate #int sqrt(5+4x-x^2)# by trigonometric substitution?

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Answer 1 · 2016-11-25T06:28:01

The answer is #=9/2arcsin((x-2)/3)+(x-2)/2sqrt(9-(x-2)^2)+C#

Let's complete the square

#5+4x-x^2=9-(2-x)^2#

Let #x-2=3sinu#

#dx=3cosudu#

#9-(2-x)^2=9-9sin^2u=9cos^2u#

#intsqrt(5+4x-x^2)dx=int3cosu*3cosudu#

#=9intcos^2udu#

#cos2u=2cos^2u-1#

#cos^2u=(1+cos2u)/2#

#9intcos^2udu=9/2int(1+cos2u)du#

#=9/2(u+(sin2u)/2)=9/2(u+sinucosu)#

#sinu=(x-2)/3#

#u=arcsin((x-2)/3)#

#cosu=1/3sqrt(9-(x-2)^2)#

#9/2(u+sinucosu)=9/2(arcsin((x-2)/3)+(x-2)/9sqrt(9-(x-2)^2))#

#intsqrt(5+4x-x^2)dx=9/2arcsin((x-2)/3)+(x-2)/2sqrt(9-(x-2)^2)+C#