How do you integrate #int 1/(xsqrt(16x^2-9))# by trigonometric substitution?

1 Answer
Mar 18, 2018

#int dx/(xsqrt(16x^2-9)) = x/(3absx) arcsec(4/3x)+C#

Explanation:

#int dx/(xsqrt(16x^2-9)) = int dx/(3xsqrt ((4/3x)^2 -1))#

Substitute:

#4/3 x = sect#, #dx = 3/4 sect tant dt#

and restrict for now to #x in (3/4,+oo)# so that #t in (0,pi/2)#

#int dx/(xsqrt(16x^2-9)) = 3/4 int (sect tant dt)/(9/4 sect sqrt(sec^2t-1))#

#int dx/(xsqrt(16x^2-9)) = 1/3 int ( tant dt)/ sqrt(sec^2t-1)#

Use now the trigonometric identity:

#sec^2t -1 = 1/cos^2t -1 = (1-sin^2t)/cos^2t = tan^2t#

As #t in (0,pi/2) => tant >=0#

then:

#sqrt(sec^2t-1) = tant#

and:

#int dx/(xsqrt(16x^2-9)) = 1/3 int ( tant dt)/ tant = 1/3 int dt =t/3 +C#

and undoing the substitution:

#int dx/(xsqrt(16x^2-9)) = 1/3 arcsec(4/3x)+C#

For #x in(-oo,-3/4)# we can use the same procedure except that:

#sqrt(sec^2t-1) = -tant#

and accordingly:

#int dx/(xsqrt(16x^2-9)) = -1/3 arcsec(4/3x)+C#

So we can express the integral for any #x# as:

#int dx/(xsqrt(16x^2-9)) = x/(3absx) arcsec(4/3x)+C#

Note that as:

#arcsec x = arctan (xsqrt((x^2-1)/x^2))#

We can also express the integral as:

#int dx/(xsqrt(16x^2-9)) = x/(3absx) arctan ((4x)/3 sqrt (( 16x^2-9)/(16x^2))) +C#

#int dx/(xsqrt(16x^2-9)) = x/(3absx) arctan ( sqrt ( 16x^2-9)/3) +C#