int dx/(xsqrt(16x^2-9)) = int dx/(3xsqrt ((4/3x)^2 -1))∫dxx√16x2−9=∫dx3x√(43x)2−1
Substitute:
4/3 x = sect43x=sect, dx = 3/4 sect tant dtdx=34secttantdt
and restrict for now to x in (3/4,+oo)x∈(34,+∞) so that t in (0,pi/2)t∈(0,π2)
int dx/(xsqrt(16x^2-9)) = 3/4 int (sect tant dt)/(9/4 sect sqrt(sec^2t-1))∫dxx√16x2−9=34∫secttantdt94sect√sec2t−1
int dx/(xsqrt(16x^2-9)) = 1/3 int ( tant dt)/ sqrt(sec^2t-1)∫dxx√16x2−9=13∫tantdt√sec2t−1
Use now the trigonometric identity:
sec^2t -1 = 1/cos^2t -1 = (1-sin^2t)/cos^2t = tan^2tsec2t−1=1cos2t−1=1−sin2tcos2t=tan2t
As t in (0,pi/2) => tant >=0t∈(0,π2)⇒tant≥0
then:
sqrt(sec^2t-1) = tant√sec2t−1=tant
and:
int dx/(xsqrt(16x^2-9)) = 1/3 int ( tant dt)/ tant = 1/3 int dt =t/3 +C∫dxx√16x2−9=13∫tantdttant=13∫dt=t3+C
and undoing the substitution:
int dx/(xsqrt(16x^2-9)) = 1/3 arcsec(4/3x)+C∫dxx√16x2−9=13arcsec(43x)+C
For x in(-oo,-3/4)x∈(−∞,−34) we can use the same procedure except that:
sqrt(sec^2t-1) = -tant√sec2t−1=−tant
and accordingly:
int dx/(xsqrt(16x^2-9)) = -1/3 arcsec(4/3x)+C∫dxx√16x2−9=−13arcsec(43x)+C
So we can express the integral for any xx as:
int dx/(xsqrt(16x^2-9)) = x/(3absx) arcsec(4/3x)+C∫dxx√16x2−9=x3|x|arcsec(43x)+C
Note that as:
arcsec x = arctan (xsqrt((x^2-1)/x^2))arcsecx=arctan(x√x2−1x2)
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∫dxx√16x2−9=x3|x|arctan(4x3√16x2−916x2)+C
int dx/(xsqrt(16x^2-9)) = x/(3absx) arctan ( sqrt ( 16x^2-9)/3) +C∫dxx√16x2−9=x3|x|arctan(√16x2−93)+C
