Integral

Calculus (Integrals, Series)
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Integral

#1

Post by Tolaso J Kos »

Prove that: $$\int_{0}^{1/\sqrt{2}}\frac{\sqrt{1+x^4}}{1-x^4}\,dx=\frac{1}{2\sqrt{2}}\left ( \frac{\pi}{2}+\ln \left ( 2+\sqrt{5} \right )-\tan^{-1}\frac{\sqrt{5}}{2} \right )$$
Imagination is much more important than knowledge.
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Integral

#2

Post by Papapetros Vaggelis »

Hi Tolis.

The function \(\displaystyle{f:\left[0,\dfrac{1}{\sqrt{2}}\right]\longrightarrow \mathbb{R}}\) is well defined and continuous at

\(\displaystyle{\left[0,\dfrac{1}{\sqrt{2}}\right]}\), so : \(\displaystyle{\int_{0}^{\frac{1}{\sqrt{2}}}f(x)\,\mathrm{d}x<\infty}\).

Consider the indefinite integral \(\displaystyle{I=\int \dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x}\) at \(\displaystyle{\left[0,\dfrac{1}{\sqrt{2}}\right]}\).

\(\displaystyle{\int \dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x=\int \dfrac{\sqrt{1+x^4}}{\left(1-x^2\right)\,\left(1+x^2\right)}\,\mathrm{d}x}\).

By substituting \(\displaystyle{t=\arctan\,x\,,t\in\left[0,\arctan\,\dfrac{1}{\sqrt{2}}\right]\subseteq \left[0,\dfrac{\pi}{4}\right)}\), we get :

\(\displaystyle{\mathrm{d}t=\dfrac{1}{1+x^2}\,\mathrm{d}x}\) and \(\displaystyle{x=\tan\,t}\), so :

\(\displaystyle{\begin{aligned}I&=\int \dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x\\&=\int \dfrac{\sqrt{1+\tan^4\,t}}{1-\tan^2\,t}\,\mathrm{d}t\\&=\int \dfrac{\displaystyle{\sqrt{\dfrac{\sin^4\,t+\cos^4\,t}{\cos^4\,t}}}}{\displaystyle{\dfrac{\cos^2\,t-\sin^2\,t}{\cos^2\,t}}}\,\mathrm{d}t\\&=\int \dfrac{\sqrt{\left(\sin^2\,t+\cos^2\,t\right)^2-2\,\sin^2\,t\,\cos^2\,t}}{\cos\,(2\,t)}\,\mathrm{d}t\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\sqrt{2-\sin^2\,(2\,t)}}{1-\sin^2\,(2\,t)}\,2\,\cos\,(2\,t)\,\mathrm{d}t\end{aligned}}\)

Now, setting \(\displaystyle{u=\sin\,(2\,t)\,,u\in\left[0,1\right)}\), we have that : \(\displaystyle{\mathrm{d}u=2\,\cos\,(2\,t)\,\mathrm{d}t}\) and thus :

\(\displaystyle{I=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\sqrt{2-u^2}}{1-u^2}\,\mathrm{d}u}\).

By applying the substitution \(\displaystyle{u=\sqrt{2}\,\sin\,y\,,y\in\left[0,\dfrac{\pi}{4}\right)}\), we get : \(\displaystyle{\mathrm{d}u=\sqrt{2}\,\cos\,y\,\mathrm{d}y}\) and then :

\(\displaystyle{\begin{aligned} I&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\sqrt{2-2\,\sin^2\,y}}{1-2\,\sin^2\,y}\,\sqrt{2}\,\cos\,y\,\mathrm{d}y\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{2\,\cos^2\,y}{\cos\,(2\,y)}\,\mathrm{d}y\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\cos\,(2\,y)+1}{\cos\,(2\,y)}\,\mathrm{d}y\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \left(1+\dfrac{1}{\cos\,(2\,y)}\right)\,\mathrm{d}y\end{aligned}}\)

where \(\displaystyle{\int 1\,\mathrm{d}y=y+c\,,c\in\mathbb{R}}\) and

\(\displaystyle{\begin{aligned} \int \dfrac{1}{\cos\,(2\,y)}\,\mathrm{d}y&=\int \dfrac{\cos\,(2\,y)}{\cos^2\,(2\,y)}\,\mathrm{d}y\\&=\dfrac{1}{2}\,\int \dfrac{1}{1-\sin^2\,(2\,y)}\,\mathrm{d}(\sin\,(2\,y))\\&=\dfrac{1}{4}\,\int \dfrac{\left(1-\sin\,(2\,y)\right)+\left(1+\sin\,(2\,y)\right)}{\left(1-\sin\,(2\,y)\right)\,\left(1+\sin\,(2\,y)\right)}\,\mathrm{d}(\sin\,(2\,y))\\&=\dfrac{1}{4}\,\int \left[\dfrac{1}{1+\sin\,(2\,y)}+\dfrac{1}{1-\sin\,(2\,y)}\right]\,\mathrm{d}(\sin\,(2\,y))\\&=\dfrac{1}{4}\,\ln\,\left|\dfrac{1+\sin\,(2\,y)}{1-\sin\,(2\,y)}\right|+c'\,,c'\in\mathbb{R}\end{aligned}}\)

so :

\(\displaystyle{I=y+\dfrac{1}{4}\,\ln\,\left|\dfrac{1+\sin\,(2\,y)}{1-\sin\,(2\,y)}\right|+c\,,c\in\mathbb{R}}\).

Now, we have that :

\(\displaystyle{u=\sqrt{2}\,\sin\,y \implies u^2=2\,\sin^2\,y=1-\cos\,(2\,y)\implies u^2-1=-\cos\,(2\,y)\implies \left(u^2-1\right)^2=\cos^2\,(2\,y)=1-\sin^2\,(2\,y)}\)

or : \(\displaystyle{\sin^2\,(2\,y)=1-\left(u^2-1\right)^2\iff \sin\,(2\,y)=\sqrt{1-\left(u^2-1\right)^2}=\sqrt{u^2\,\left(2-u^2\right)}=u\,\sqrt{2-u^2}}\).

Also, \(\displaystyle{y=\arcsin\,\dfrac{u}{\sqrt{2}}=\arcsin\,\dfrac{\sin\,(2\,t)}{\sqrt{2}}=\arcsin\,\dfrac{\sin\,(2\,\arctan\,x)}{\sqrt{2}}}\).

\(\displaystyle{u=\sin\,(2\,t)=\sin\,(2\,\arctan\,x)}\), so :

\(\displaystyle{u^2-1=\sin^2\,(2\,\arctan\,x)-1=-\cos^2\,(2\,\arctan\,x)}\) and then :

\(\displaystyle{\sin\,(2\,y)=\sqrt{1-\left(u^2-1\right)^2}=\sqrt{1-\cos^4\,(2\,\arctan\,x)}=\sin\,(2\,\arctan\,x)\,\sqrt{1+\cos^2\,(2\,\arctan\,x)}}\).

In conclusion : $$ I=\frac{1}{2\sqrt{2}}\left [ \arcsin\frac{\sin \left ( 2\arctan x \right ) }{\sqrt{2}} +\frac{1}{4}\ln \left | \frac{1+\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2 \left ( 2\arctan x \right )}}{1-\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2\left ( 2\arctan x \right )}} \right | \right ]+c, \; c \in \mathbb{R}$$ and those functions are continuous and differentiable at \(\displaystyle{\left[0,\dfrac{1}{\sqrt{2}}\right]}\). Then :

\(\displaystyle{\int_{0}^{\frac{1}{\sqrt{2}}}\,\dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x=}\) $$=\left [ \frac{1}{2\sqrt{2}}\left [ \arcsin\frac{\sin \left ( 2\arctan x \right ) }{\sqrt{2}} +\frac{1}{4}\ln \left | \frac{1+\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2 \left ( 2\arctan x \right )}}{1-\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2\left ( 2\arctan x \right )}} \right | \right ] \right ] _0^{1/\sqrt{2}}$$

where :

\(\displaystyle{\sin\,(2\,\arctan\,x)=2\,\sin\,(\arctan\,x)\,\cos\,(\arctan\,x)}\) and

\(\displaystyle{\cos\,(\arctan\,x)=\dfrac{1}{\sqrt{1+\tan^2\,(\arctan\,x)}}=\dfrac{1}{\sqrt{1+x^2}}}\) ,

and then, since \(\displaystyle{\sin\,(\arctan\,x)>0\,,0\leq x\leq \dfrac{1}{\sqrt{2}}} \), we have that :

\(\displaystyle{\sin\,(\arctan\,x)=\sqrt{1-\cos^2\,(\arctan\,x)}=\dfrac{x}{\sqrt{1+x^2}}}\).

\(\displaystyle{I=\dfrac{1}{2\,\sqrt{2}}\,\left(f_{c}\left(\dfrac{1}{\sqrt{2}}\right)-f_{c}(0)\right)}\) gives the desired result.
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Re: Integral

#3

Post by Tolaso J Kos »

Good evening Vaggelis..
Here's another approach: $$\begin{aligned}
\int_{0}^{1/\sqrt{2}}\frac{\sqrt{1+x^4}}{1-x^4}\,dx&\overset{x=\sqrt{\tan u}}{=\! =\! =\! =\! =\!}\int_{0}^{\tan^{-1}1/2}\frac{\sec u }{\sec^2 u \cos 2u}\frac{\sec^2 u}{2\sqrt{\tan u}},du\\
&= \frac{1}{2}\int_{0}^{\tan^{-1}1/2}\frac{du}{\cos 2u \sqrt{\frac{\sin 2u}{2}}}\\
&=\frac{\sqrt{2}}{2}\int_{0}^{\tan^{-1}1/2} \frac{\cos 2u}{\left ( 1-\sin^2 2u \right )\sqrt{\sin 2u}}\,du \\
&=\frac{\sqrt{2}}{2}\int_{0}^{2/\sqrt{5}}\frac{ds}{1-s^4}\\
&=\cdots \\
&= \frac{1}{2\sqrt{2}}\left ( \frac{\pi}{2}+\ln \left ( 2+\sqrt{5}\right)-\tan^{-1}\frac{\sqrt{5}}{2} \right )
\end{aligned}$$
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 9 guests