How do you integrate int 1/sqrt(4x^2+16x+8) 14x2+16x+8 using trigonometric substitution?

2 Answers
Jun 16, 2018

1/2ln|x+2+sqrt(x^2+4x+2)|+C12lnx+2+x2+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)dx14x2+16x+8dx=121x2+4x+2dx

Complete the square:

1/2int1/sqrt((x+2)^2-2)dx121(x+2)22dx

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

As x=-2+sqrt(2)secthetax=2+2secθ, dx/(d theta)=sqrt(2)secthetatanthetadxdθ=2secθtanθ, and so our integral becomes

1/2int1/sqrt(2sec^2theta-2)*sqrt(2)secthetatanthetad theta1212sec2θ22secθtanθdθ

Recalling the identity tan^2theta=sec^2theta-1tan2θ=sec2θ1, we obtain

1/2int1/(sqrt(2)tantheta)*sqrt(2)secthetatanthetad theta1212tanθ2secθtanθdθ

which simplifies considerably to become

1/2intsectheta d theta12secθdθ

The integral of secxsecx is ln|secx+tanx|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|+C12ln|secθ+tanθ|+C

Transform back to our original variable, xx:

1/2ln|(x+2)/sqrt(2)+sqrt((x+2)^2/2-1)|+C12ln∣ ∣x+22+(x+2)221∣ ∣+C

Rearrange a little to obtain a simpler form:

1/2ln|(x+2)/sqrt(2)+sqrt(((x+2)^2-2)/2)|+C12ln∣ ∣x+22+(x+2)222∣ ∣+C
1/2ln|x+2+sqrt((x+2)^2-2)|-1/2lnsqrt(2)+C12lnx+2+(x+2)2212ln2+C

Note that the -1/2lnsqrt(2)12ln2 term is a constant and can therefore be folded into the integration constant CC

1/2ln|x+2+sqrt(x^2+4x+2)|+C12lnx+2+x2+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 1/2ln212ln2 without changing the answer:

1/2ln|x+2+sqrt(x^2+4x+2)|+C12lnx+2+x2+4x+2+C
1/2ln|x+2+sqrt(x^2+4x+2)|+1/2ln2+C12lnx+2+x2+4x+2+12ln2+C
1/2ln|2(x+2)+sqrt(4x^2+16x+8)|+C12ln2(x+2)+4x2+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)|+C12lnx+2+x2+4x+2+C

It's more elegant.

Jun 16, 2018

I=1/2ln|(x+2)+sqrt(x^2+4x+2)|+CI=12ln(x+2)+x2+4x+2+C

Explanation:

Here,

I=int1/sqrt(4x^2+16x+8)dxI=14x2+16x+8dx

I=int1/(2sqrt(x^2 +4x+2))dxI=12x2+4x+2dx

I=1/2int1/sqrt(x^2+4x+4-2)dxI=121x2+4x+42dx

I=1/2int1/sqrt((x+2)^2-(sqrt2)^2)dxI=121(x+2)2(2)2dx

Subst. x+2=sqrt2secu=>dx=sqrt2secutanudux+2=2secudx=2secutanudu

=>secu=(x+2)/sqrt2secu=x+22

I=1/2int1/sqrt(2sec^2u-2)xxsqrt2secutanuduI=1212sec2u2×2secutanudu

:.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