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

1 Answer
Sep 6, 2016

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

Explanation:

We have:

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

From here, let #3sintheta=x+2#. Thus, #3costhetad theta=dx#. Plugging these in:

#I=3intcosthetasqrt(-9sin^2theta+9)d theta#

Factoring a #sqrt9=3# from the square root:

#I=9intcosthetasqrt(-sin^2theta+1)d theta#

Recall that since #sin^2theta+cos^2theta=1#, we know that #sqrt(-sin^2theta+1)=costheta#.

#I=9intcos^2thetad theta#

Recall that #cos2theta=2cos^2theta-1#. Thus, #cos^2theta=1/2(cos2theta+1)#.

#I=9/2int(cos2theta+1)d theta=9/2intcos2thetad theta+9/2intd theta#

The first integral can be approached with substitution, or just the reverse chain rule. The second is simple:

#I=9/4int2cos2thetad theta+9/2theta=9/4sin2theta+9/2theta#

Using #sin2theta=2sinthetacostheta#:

#I=9/2sinthetacostheta+9/2theta#

Writing all in terms of sine:

#I=9/2sinthetasqrt(1-sin^2theta)+9/2theta#

Recall that #sintheta=(x+2)/3#, so we also see that #theta=arcsin((x+2)/3)#:

#I=9/2((x+2)/3)sqrt(1-((x+2)/3)^2)+9/2arcsin((x+2)/3)#

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

The #sqrt(1/9)=1/3# will come out of the square root and cancel with the #3#:

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

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