How do you integrate #int (25-x^2)/sqrt(x^2+4)dx# using trigonometric substitution?

We will want to use the substitution #x=2tantheta#. This implies that #dx=2sec^2thetacolor(white).d theta#.

#I=int(25-x^2)/sqrt(x^2+4)dx=int(25-4tan^2theta)/sqrt(4tan^2theta+4)(2sec^2thetacolor(white).d theta)#

Note that #sqrt(4tan^2theta+4)=2sqrt(tan^2theta+1)=2sectheta#. The fact that we use #tan^2theta+1=sec^2theta# to simplify the square root was the reason to choose the substitution #x=2tantheta# in the first place.

#I=int(25-4tan^2theta)/(2sectheta)(2sec^2theta)d theta=int(25-4tan^2theta)(sec theta)d theta#

Expanding and writing #tan^2thetasectheta# in terms of #sintheta# and #costheta#:

#I=25intsecthetacolor(white).d theta-4intsin^2theta/cos^3thetad theta#

Rewrite #sin^2theta# as #1-cos^2theta#:

#I=25intsecthetacolor(white).d theta-4int(1-cos^2theta)/cos^3thetad theta#

#I=25intsecthetacolor(white).d theta-4int(sec^3theta-sectheta)d theta#

#I=29intsecthetacolor(white).d theta-4intsec^3thetacolor(white).d theta" "" "color(red)((star))#

We will tackle #intsec^3thetacolor(white).d theta# with integration by parts. Since #sec^3theta=sectheta(sec^2theta)#, and #sec^2theta# can easily be integrated, let:

#{(u=sectheta,=>,du=secthetatanthetacolor(white).d theta),(dv=sec^2thetacolor(white).d theta,=>,v=tantheta):}#

Then, using #intudv=uv-intvdu#:

#I=29intsecthetacolor(white).d theta-4(secthetatantheta-intsecthetatan^2thetacolor(white).d theta)#

#I=29intsecthetacolor(white).d theta-4secthetatantheta+4intsecthetatan^2thetacolor(white).d theta#

Now letting #tan^2theta=sec^2theta-1#:

#I=29intsecthetacolor(white).d theta-4secthetatantheta+4intsectheta(sec^2theta-1)d theta#

#I=25intsecthetacolor(white).d theta-4secthetatantheta+[4intsec^3thetacolor(white).d theta]#

Returning to #color(red)((star))#, we see that #4intsec^3thetacolor(white).d theta=29intsecthetacolor(white).d theta-I#:

#I=25intsecthetacolor(white).d theta-4secthetatantheta+[29intsecthetacolor(white).d theta-I]#

Combining and adding #I# to both sides:

#2I=54intsecthetacolor(white).d theta-4secthetatantheta#

#I=27intsecthetacolor(white).d theta-2secthetatantheta#

The antiderivative of #sectheta# is commonly known:

#I=27lnabs(tantheta+sectheta)-2secthetatantheta#

Our original substitution was #x=2tantheta#, which implies that #tantheta=x/2#.

From this we can say that #sectheta=sqrt(tan^2theta+1)=sqrt(x^2/4+1)=1/2sqrt(x^2+4)#. Then:

#I=27lnabs(x/2+1/2sqrt(x^2+4))-2(1/2sqrt(x^2+4))x/2+C#

Factoring #1/2# from the natural logarithm, it can be removed from the logarithm using #ln(a/b)=ln(a)-ln(b)#, and the resultant #-ln(2)# will be absorbed into the constant of integration.

#I=27lnabs(x+sqrt(x^2+4))-1/2xsqrt(x^2+4)+C#

#int frac{25-x^2}{sqrt(x^2+4)}dx#

#=int frac{25-x^2}{sqrt[(4)(x^2/4+1)]}#

#=1/2int frac{25-x^2}{sqrt((x/2)^2+1)}#

Recognize trig substitution in denominator from radical, so list various useful trigonometric ratios with this triangle.

#color(green)(tantheta=x/2)#
#color(green)(x=2tan theta)#
#color(green)(dx=2sec^2theta (d theta))#
#color(green)(25-x^2=25-4tan^2theta)#

#color(green)(sectheta=sqrt[(x/2)^2+1])#

#=1/2intfrac{25-4tan^2theta}{sectheta}(2sec^2theta)(d theta)#

#=1/2intfrac{25-4tan^2theta}{cancel(sectheta)}(2cancel(sectheta)sectheta)(d theta)#

#=int[25sectheta-4secthetatan^2theta] d theta #

#=25int(sectheta )d theta-4int(secthetatan^2theta )d theta#

#=25ln|sectheta+tantheta|-4int(secthetatan^2theta)d theta#

Now, we're getting close to the end - we have to simplify #color(red)(int(secthetatan^2theta)d theta)#
Use integration by parts:
#color(red)(u=tantheta)#
#color(red)(du=sec^2theta)#
#color(red)(v=sectheta)#
#color(red)(dv=secthetatantheta)#

#color(red)(int(secthetatan^2theta)d theta=secthetatantheta-int(sec^3theta) d theta)#
#color(red)(=secthetatantheta-int(sectheta)(1+tan^2theta)d theta)#
#color(red)(=secthetatantheta-int(sectheta)d theta-int(secthetatan^2theta)d theta)#
#color(red)(int(secthetatan^2theta)d theta=secthetatantheta-ln|sectheta+tantheta|-int(secthetatan^2theta)d theta)#
#color(red)(2intsecthetatan^2theta d theta=secthetatantheta-ln|sectheta+tantheta|)#
#color(red)(intsecthetatan^2theta d theta=1/2secthetatantheta-1/2ln|sectheta+tantheta|+C)#

#=25ln|sectheta+tantheta|-4[1/2secthetatantheta-1/2ln|sectheta+tantheta|]+C#

#=25ln|sectheta+tantheta|-2secthetatantheta+2ln|sectheta+tantheta|+C#

#=27ln|sectheta+tantheta|-2secthetatantheta+C#

Lastly, re-substitute the trig ratios:
#=27ln|sqrt[(x/2)^2+1]+x/2|-2(x/2)sqrt[(x/2)^2+1]+C#

#=27ln|x/2 + sqrt[(x/2)^2+1]|-xsqrt[(x/2)^2+1]+C#

Change the square root notation back to the original form:
#=27ln|[x+sqrt[x^2+4]]/2|-(xsqrt[x^2+4])/2+C#