How do I integrate #intsqrt(x²-49)dx#?

1 Answer
Jul 9, 2015

#sqrt(x^2 - 49) prop sqrt(x^2 - a^2)#

Let:
#x = asectheta# with #a = 7#:
#dx = 7secthetatanthetad theta#
#sqrt(x^2 - 49) = sqrt(7^2sec^2theta - 7^2) = 7sqrt(sec^2theta - 1)#

#= 7tantheta#

Thus:

#= int 7tantheta * 7secthetatanthetad theta#

#= 49int secthetatan^2thetad theta#

#= 49int sectheta(sec^2theta - 1)d theta#

#= color(highlight)(49)int sec^3theta- secthetad theta#

Solving these individually:

Small trick:
#int secthetad theta = int sectheta((sectheta + tantheta)/(sectheta + tantheta))d theta#

#= int (sec^2theta + secthetatantheta)/(sectheta + tantheta)d theta#

u-substitution. Let:
#u = sectheta + tantheta#
#du = secthetatantheta + sec^2thetad theta#

#=> int 1/udu#

#= ln|u| = color(green)(ln|sectheta + tantheta|)#

The other one:

#int sec^3thetad theta#

Integration by Parts. Let:
#u = sectheta#
#du = secthetatantheta#
#dv = sec^2thetad theta#
#v = tantheta#

#=> uv - int vdu#

#= secthetatantheta - int secthetatan^2thetad theta#

#= secthetatantheta - int sec^3theta - secthetad theta#

#int sec^3thetad theta = secthetatantheta - int sec^3thetad theta + int secthetad theta#

Add the #sec^3theta# over. No need to evaluate it. We also know #intsecthetad theta# already.

#2int sec^3thetad theta = secthetatantheta + int secthetad theta#

#2int sec^3thetad theta = secthetatantheta + ln|sectheta + tantheta|#

#int sec^3thetad theta = color(green)(1/2[secthetatantheta + ln|sectheta + tantheta|])#

Overall:

#= overbrace(1/2[secthetatantheta + ln|sectheta + tantheta|])^(intsec^3thetad theta) - overbrace(ln|sectheta + tantheta|)^(intsecthetad theta)#

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

Recall that #sectheta = x/7# and #tantheta = sqrt(x^2 - 49)/7#, and don't forget the #color(highlight)(49)#:

#= 49{1/2[x/7*sqrt(x^2 - 49)/7 - ln|x/7 + sqrt(x^2 - 49)/7|]}#

#= 1/2[(xsqrt(x^2 - 49)) - 49ln|x/7 + sqrt(x^2 - 49)/7|]#

#= 1/2[(xsqrt(x^2 - 49)) - 49ln|(1/7)(x + sqrt(x^2 - 49))|]#

#= 1/2[(xsqrt(x^2 - 49)) - 49(ln|x + sqrt(x^2 - 49)| + ln(1/7))] + C#

The remaining #-49/2ln(1/7)# gets embedded into #C#:

#= color(blue)(1/2[xsqrt(x^2 - 49) - 49ln|x + sqrt(x^2 - 49)|] + C)#