Compute $\int \sqrt{1+4x^2} \, dx$ with Euler substitution

Question

In this post: Computing $\int \sqrt{1+4x^2} \, dx$ someone mentioned Euler substitution to compute the following integral:

$$\int \sqrt{1+4x^2} \, dx$$

I tried to follow this advice and got very nice result, namely I substituted $\sqrt{1+4x^2}=t-2x$ which after raising to the square and reducing gives $x=\frac{t^2-1}{4t}$ and $t-2x=\frac{t^2+1}{2t}$, then derivative is equal to $\frac{dx}{dt}=\frac{t^2+1}{4t^2}$ and the whole integral:

$$\int \sqrt{1+4x^2} \, dx = \int \frac{(t^2+1)^2}{8t^3} \, dt$$

Could you please check my solution, because it seems a lot easier than all these trigonometric substitution (too easy which is suspicious...). Thanks in advance.

Answer 1

If you set $\sqrt{1+4x^2}=t-2x$, you have $$ 1+4x^2=t^2-4tx+4x^2 $$ so $4tx=t^2-1$ and therefore $$ x=\frac{t^2-1}{4t}=\frac{t}{4}-\frac{1}{4t} $$ Thus $$ dx=\left(\frac{1}{4}+\frac{1}{4t^2}\right)\,dt=\frac{t^2+1}{4t^2}\,dt $$ and $$ \sqrt{1+4x^2}=t-\frac{t}{2}+\frac{1}{2t}=\frac{t^2+1}{2t} $$ so the integral becomes $$ \int\frac{(t^2+1)^2}{8t^3}\,dt= \frac{1}{8}\int\left(t+\frac{2}{t}+\frac{1}{t^3}\right)\,dt $$ that's elementary.

Yes, your computation is right.

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The alternative way is setting $2x=\sinh(t/2)$, so $$ \sqrt{1+4x^2}=\cosh\frac{t}{2}, \qquad dx=\frac{1}{4}\cosh\frac{t}{2}\,dt $$ and the integral is $$ \frac{1}{4}\int\cosh^2\frac{t}{2}\,dt= \frac{1}{8}\int(\cosh t-1)\,dt $$ remembering that $2\cosh^2\frac{t}{2}+1=\cosh t$.

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Another (tricky) way is to do $2x=t$, that reduces to computing, up to a scalar factor, $$\DeclareMathOperator{\arsinh}{arsinh} I=\int\sqrt{1+t^2}\,dt=\int\frac{1+t^2}{\sqrt{1+t^2}}\,dt= \arsinh t+\int t\frac{t}{\sqrt{1+t^2}}\,dt= \arsinh t+t\sqrt{1+t^2}-I $$ (the last one by parts).

Answer 2

Hint $$x=\frac{1}{2}\tan\theta$$ $$I=\frac12\int\sec^3\theta\,d\theta $$