How do you integrate int 1/sqrt(4x^2-12x-7) using trigonometric substitution?

2 Answers
Jun 27, 2018

(2x-a)^2=4x^2-4ax+a^2

Explanation:

int1/sqrt(4x^2-12x-7)dx=int1/sqrt((2x-3)^2-16)dx

int1/sqrt(16((2x-3)^2/16-1))dx=1/4int1/sqrt(((2x-3)/4)^2-1)dx

Let u=x/2-3/4 therefore 2du=dx

1/2int1/sqrt(u^2-1)=i/2int-1/sqrt(1-u^2)

i/2cos^-1((2x-3)/4)

Jun 27, 2018

The answer is =1/2ln(|(2x-3)/4+sqrt((((2x-3)/(4))^2)-1)|)+C

Explanation:

Rewrite

4x^2-12x-7=(2x-3)^2-16

=4^2(((2x-3)/4)^2-1)

Therefore,

sqrt(4x^2-12x-7)=sqrt(4^2(((2x-3)/4)^2-1))

=4sqrt((((2x-3)/4)^2-1))

The integral is

I=int(dx)/(sqrt(4x^2-12x-7))=1/4int(dx)/(sqrt((((2x-3)/4)^2-1)))

Let u=(2x-3)/4, =>, du=dx/2

Therefore,

I=1/4int(2du)/(sqrt(u^2-1))=1/2int(du)/sqrt(u^2-1)

Let u=secv, =>, du=secvtanvdv

sec^2v-1=tan^2v

The integral is

I=1/2int(secvtanvdv)/sqrt(sec^2v-1)

=1/2intsecvdv

=1/2int(secv(secv+tanv)dv)/(secv+tanv)

Let w=secv+tanv, =>, dw=(secvtanv+sec^2v)dv

Therefore,

I=1/2int(dw)/w

=1/2ln(w)

=1/2ln(secv+tanv)

=1/2ln(u+sqrt(u^2-1))

=1/2ln((2x-3)/4+sqrt((((2x-3)/(4))^2)-1))

=1/2ln(|(2x-3)/4+sqrt((((2x-3)/(4))^2)-1)|)+C