How do I solve the following $\int_0^1\frac1{\sqrt{1+x^4}}\,\mathrm dx$

Question

I'm trying to solve the following integral: $$\int_0^1 \frac{1}{\sqrt{1+x^4}} \,dx$$

Here's a link to the original question:

https://photos.app.goo.gl/mfTe649pzgKHeUKy7

I've tried to substitute $x$ as $\tan t$ and then $2t$ as $u$. Eventually I end up with:

\begin{align*} \int_0^{\pi / 2} (\cos u)^{-1/2} \, du \end{align*}

Answer 1

On the interval $[0,1]$ $$1\geq1-x^8$$ Therefore $$\frac{1}{1+x^4}\geq1-x^4=\left(1+x^2\right)\left(1-x^2\right)\geq1-x^2$$ Therefore $$\frac{1}{\sqrt{1+x^4}}\geq\sqrt{1-x^2}$$ Therefore $$\int_0^1\frac{1}{\sqrt{1+x^4}}dx\geq\int_0^1\sqrt{1-x^2}dx=\frac{\pi}{4}>\frac34$$ Also, it is clear the integral in question is less than $1$, since $\frac{1}{\sqrt{1+x^4}}\leq1$, and therefore $$\int_0^1\frac{1}{\sqrt{1+x^4}}dx\leq\int_0^11\,dx=1$$

So $\int_0^1\frac{1}{\sqrt{1+x^4}}dx$ is in $\left[\frac34,1\right]$. There is no smaller interval among the options listed in the multiple choice version of the question that is linked.

Answer 2

You don't need to actually evaluate the integral to answer your question.

It is clear that $\frac{1}{\sqrt{1+x^4}} ≤ 1$ since $x^4 ≥ 0 \Rightarrow 1+x^4 ≥ 1$.

For all real $x$ in the interval $[0,1]$, $1 + x^4 ≤ 1+x^2$, so $\frac{1}{\sqrt{1+x^4}} ≥ \frac{1}{\sqrt{1+x^2}}.$

We can use the fact that $\int \frac{1}{\sqrt{1+x^2}} \ \mathrm d x = \sinh^{-1} x $, so: $$\int_0^1 \frac{1}{\sqrt{1+x^2}} \mathrm d x = \left[\sinh^{-1} x\right]_0^1 = \sinh^{-1} 1 \approx 0.881.$$

Therefore the value of the integral lies in $\left[\frac{3}{4}, 1 \right]$, so option $4$ is correct.

---

Here is a proof that $\int \frac{1}{\sqrt{1+x^2}} \ \mathrm d x = \sinh^{-1} x $:

Substitute $x = \tan u, \mathrm d x = \sec^2 u \ \mathrm d u$:

$$\int \frac{1}{\sqrt{1+\tan^2 u}} \ \sec^2 u \ \mathrm d u$$ $$=\int \sec u \ \mathrm d u$$ $$=\ln(\tan u + \sec u)$$

Now draw a right triangle with legs $1, \tan u$ and hypotenuse $\sqrt{1+\tan^2 u}$. We find that $\sec u = \frac{\sqrt{1+\tan^2 u}}{1} = \sqrt{1+\tan^2 u} = \sqrt{1+x^2}$.

We thus have: $$=\ln(x + \sqrt{1+x^2})$$ $$=\sinh^{-1} x$$

according to the definition of $\sinh^{-1} x$. This can also be shown by letting $y = \sinh x = \frac{e^x+e^{-x}}{2}$, solving for $x$ using the quadratic formula then rearranging (the proof of the last step is widely available online such as Yahoo Answers).

Answer 3

Moving on from your last integral, there exists a function:

$$\beta(x,y) := 2\int_0^{\pi/2} \sin^{2x-1}(x)\cos^{2y-1}dx$$

So, with the exponent over the cosine being -1/2 and the exponent over the sine being technically 0, we can conclude $x$ must be $\frac{1}{4}$ and $y$ must be $\frac{1}{2}$. Thus, if we let your integral be $I$,

$$I = \frac{1}{2}\beta(\frac{1}{4},\frac{1}{2})$$

An identity exists such that:

$$\beta(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Therefore,

$$I = \frac{\Gamma\bigl(\frac{1}{4}\bigr)\Gamma\bigl(\frac{1}{2}\bigr)}{2\Gamma\bigl(\frac{3}{4}\bigr)}$$

Note though, I typed the integral with the reciprocal of the square root of the cuartic into wolfram alpha and it did not match that answer. So maybe you made an error in the substitution or I’m missing something in your work?