How do you integrate int 1/sqrt(4x^2+16x+8) using trigonometric substitution?

2 Answers
Jun 16, 2018

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

Explanation:

It is in general a good idea to divide any polynomial that we have to work with through by its leading coefficient; this almost always makes it simpler to work with. So:

int1/sqrt(4x^2+16x+8)dx=1/2int1/sqrt(x^2+4x+2)dx

Complete the square:

1/2int1/sqrt((x+2)^2-2)dx

Make a substitution to cause the square root to vanish via a trig identity. In this case we'll use x+2=sqrt(2)sectheta. We could also use csc (which would be similar) or, straying into hyperbolic functions, cosh (which would be easier).

As x=-2+sqrt(2)sectheta, dx/(d theta)=sqrt(2)secthetatantheta, and so our integral becomes

1/2int1/sqrt(2sec^2theta-2)*sqrt(2)secthetatanthetad theta

Recalling the identity tan^2theta=sec^2theta-1, we obtain

1/2int1/(sqrt(2)tantheta)*sqrt(2)secthetatanthetad theta

which simplifies considerably to become

1/2intsectheta d theta

The integral of secx is ln|secx+tanx|. Two demonstrations of this are given here: https://www.math.ubc.ca/~feldman/m121/secx.pdf
So we get

1/2ln|sectheta+tantheta|+C

Transform back to our original variable, x:

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

Rearrange a little to obtain a simpler form:

1/2ln|(x+2)/sqrt(2)+sqrt(((x+2)^2-2)/2)|+C
1/2ln|x+2+sqrt((x+2)^2-2)|-1/2lnsqrt(2)+C

Note that the -1/2lnsqrt(2) term is a constant and can therefore be folded into the integration constant C

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

If we wished to write this in the same style as the original question formula, we could use the same observation about the integration constant to add 1/2ln2 without changing the answer:

1/2ln|x+2+sqrt(x^2+4x+2)|+C
1/2ln|x+2+sqrt(x^2+4x+2)|+1/2ln2+C
1/2ln|2(x+2)+sqrt(4x^2+16x+8)|+C

It's arguable whether it's better to write it so that the expression inside the root matches that in the original integral or so that it is a simpler expression. I think I prefer our first answer myself:

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

It's more elegant.

Jun 16, 2018

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

Explanation:

Here,

I=int1/sqrt(4x^2+16x+8)dx

I=int1/(2sqrt(x^2 +4x+2))dx

I=1/2int1/sqrt(x^2+4x+4-2)dx

I=1/2int1/sqrt((x+2)^2-(sqrt2)^2)dx

Subst. x+2=sqrt2secu=>dx=sqrt2secutanudu

=>secu=(x+2)/sqrt2

I=1/2int1/sqrt(2sec^2u-2)xxsqrt2secutanudu

:.I=1/2int(sqrt2secutanu)/(sqrt2tanu)du

=>=1/2intsecudu

:.I=1/2ln|secu+tanu|+c

=1/2ln|secu+sqrt(sec^2u-1)|+c

Subst. back , secu=(x+2)/sqrt2

I=1/2ln|(x+2)/sqrt2+sqrt((x+2)^2/2-1)|+c

I=1/2ln|(x+2)/sqrt2+sqrt((x+2)^2-2)/sqrt2|+c

I=1/2ln|((x+2)+sqrt(x^2+4x+2))/sqrt2|+c

I=1/2ln|(x+2)+sqrt(x^2+4x+2)|-1/2lnsqrt2+c

I=1/2ln|(x+2)+sqrt(x^2+4x+2)|+C,
where,C=c-1/2lnsqrt2