How do you integrate #int sqrt(9-x^2)# using trig substitutions?

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Answer 1 · 2016-08-01T19:32:04

#=1/2[9sin^(-1)(x/3) + xsqrt(9-x^2)]+C#

Let #x = 3sin(u) implies dx = 3cos(u)du #

#x = 3sin(u) implies u = sin^(-1)(x/3)#

#int sqrt(9-x^2)dx = 3int sqrt(9 - 9sin^2(u))cos(u)du#

#= 9int sqrt(1-sin^2(u))cos(u)du = 3int sqrt(cos^2(u))cos(u)du#

#=9int cos^2(u)du#

From double angle formula

#cos2theta = 2cos^2theta - 1 implies cos^2theta = 1/2(1+cos2theta)#

#=9/2int (1 + cos(2u))du#

#= 9/2[u + 1/2sin(2u)] + C#

Apply double angle formula:

#=9/2[u + sin(u)cos(u)] + C#

Rewrite #cos(u) = sqrt(1-sin^2(u))#

#=9/2[u+sin(u)sqrt(1-sin^2(u))]+C#

#=9/2[sin^(-1)(x/3) + x/3sqrt(1-(x/3)^2)]+C#

#=9/2[sin^(-1)(x/3) + x/9sqrt(9-x^2)]+C#

#=1/2[9sin^(-1)(x/3) + xsqrt(9-x^2)]+C#

Answer 2 · 2016-08-01T19:40:25

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

Let us subst. #x=3sint#, so, #dx=3costdt#

Also, #sqrt(9-x^2)=sqrt(9-9sin^2t)=sqrt(9cos^2t)=3cost#.

Hence, #I=intsqrt(9-x^2)dx#

#=int3cost*3costdt=9intcos^2tdt=9/2int(1+cos2t)dt#

#=9/2{t+sin(2t)/2}=9/2{t+sintcost}#

Here, #x=3sint rArr t=arcsin(x/3)#.

Therefore, #I=9/2{arcsin(x/3)+x/3*sqrt(9-x^2)/3}#

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