Monday Math 143

Find $I=\int_{-1}^1\frac{\sqrt{\sqrt2-\sqrt{1+x}}}{\left(2^{3/4}+\sqrt{\sqrt2+\sqrt{1+x}}\right)\sqrt{1-x^2}}\,dx$ .

The presence of the $\sqrt{1-x^2}$ term leads us to attempt a trigonometric substitution of the form $x=\sin\theta$ or $x=\cos\theta$ . For reasons that will become clear in a moment, we will choose the latter. The limits x=-1 and x=1 become θ=π and θ=0, respectively, while $dx=-\sin\theta\,d\theta$ , so we have
$\begin{eqnarray}I&=&\int_{-1}^1\frac{\sqrt{\sqrt2-\sqrt{1+x}}}{\left(2^{3/4}+\sqrt{\sqrt2+\sqrt{1+x}}\right)\sqrt{1-x^2}}\,dx\\&=&\int_{\pi}^0\frac{\sqrt{\sqrt2-\sqrt{1+\cos\theta}}}{\left(2^{3/4}+\sqrt{\sqrt2+\sqrt{1+\cos\theta}}\right)\sqrt{1-\cos^2\theta}}\left(-\sin\theta\right)\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{\sqrt2-\sqrt{1+\cos\theta}}\sin\theta}{\left(2^{3/4}+\sqrt{\sqrt2+\sqrt{1+\cos\theta}}\right)\sqrt{\sin^2\theta}}\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{\sqrt2-\sqrt{1+\cos\theta}}}{2^{3/4}+\sqrt{\sqrt2+\sqrt{1+\cos\theta}}}\,d\theta\end{eqnarray}$ .

Now, the key is to note the trigonometric identity $1+\cos\theta=2\cos^2\frac{\theta}2$ , so we see
$\begin{eqnarray}I&=&\int_0^{\pi}\frac{\sqrt{\sqrt2-\sqrt{1+\cos\theta}}}{2^{3/4}+\sqrt{\sqrt2+\sqrt{1+\cos\theta}}}\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{\sqrt2-\sqrt{2\cos^2\frac{\theta}2}}}{2^{3/4}+\sqrt{\sqrt2+\sqrt{2\cos^2\frac{\theta}2}}}\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{\sqrt2-\sqrt{2}\left|\cos\frac{\theta}2\right|}}{2^{3/4}+\sqrt{\sqrt2+\sqrt{2}\left|\cos\frac{\theta}2\right|}}\,d\theta\\&=&\int_0^{\pi}\frac{2^{1/4}\sqrt{1-\left|\cos\frac{\theta}2\right|}}{2^{3/4}+2^{1/4}\sqrt{1+\left|\cos\frac{\theta}2\right|}}\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{1-\left|\cos\frac{\theta}2\right|}}{\sqrt2+\sqrt{1+\left|\cos\frac{\theta}2\right|}}\,d\theta\end{eqnarray}$ .
Now, since we are integrating from zero to π, $\cos\frac{\theta}2$ is positive for the entire interval, and so we may remove the absolute values:
$I=\int_0^{\pi}\frac{\sqrt{1-\cos\frac{\theta}2}}{\sqrt2+\sqrt{1+\cos\frac{\theta}2}}\,d\theta$ .
Next, we can use $1+\cos\theta=2\cos^2\frac{\theta}2$ again via θ→θ/2:
$1+\cos\frac{\theta}2=2\cos^2\frac{\theta}4$ ,
and since
$1-\cos\theta=2\sin^2\frac{\theta}2$
then
$1-\cos\frac{\theta}2=2\sin^2\frac{\theta}4$ .
Plugging these in,
$\begin{eqnarray}I&=&\int_0^{\pi}\frac{\sqrt{1-\cos\frac{\theta}2}}{\sqrt2+\sqrt{1+\cos\frac{\theta}2}}\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{2\sin^2\frac{\theta}4}}{\sqrt2+\sqrt{2\cos^2\frac{\theta}4}}\,d\theta\\&=&\int_0^{\pi}\frac{\sqrt{2}\left|\sin\frac{\theta}4\right|}{\sqrt2+\sqrt{2}\left|\cos\frac{\theta}4\right|}\,d\theta\\I&=&\int_0^{\pi}\frac{\left|\sin\frac{\theta}4\right|}{1+\left|\cos\frac{\theta}4\right|}\,d\theta\end{eqnarray}$ ,
and since $\sin\frac{\theta}4$ and $\cos\frac{\theta}4$ are both positive for our entire region of integration, we have
$I=\int_0^{\pi}\frac{\sin\frac{\theta}4}{1+\cos\frac{\theta}4}\,d\theta$ ,
and since the half-angle formula for tangent says
$\tan\frac{\alpha}2=\frac{\sin\alpha}{1+\cos\alpha}$ ,
$I=\int_0^{\pi}\tan\frac{\theta}8\,d\theta$ ,
and via $u=\frac{\theta}8$ ,
$I=8\int_0^{\pi/8}\tan{u}\,du=8\left[\ln\left|\sec{u}\right|\right]_0^{\pi/8}=8\ln\sec\frac{\pi}8$ ,

We can find an exact value for the secant above using half-angle identities along with $\sec\theta=\frac1{\cos\theta}$ and $\cos\frac{\pi}{4}=\frac1{\sqrt2}$ , we can find that
$\begin{eqnarray}\sec\frac{\pi}8&=&\frac1{\cos\frac{\pi}8}\\&=&\frac1{\sqrt{\frac{1+\cos\frac{\pi}4}2}}\\&=&\frac{\sqrt2}{\sqrt{1+\frac1{\sqrt2}}}\\&=&\frac{2^{3/4}}{\sqrt{1+\sqrt2}}\end{eqnarray}$ ,
so
$\begin{eqnarray}I&=&8\ln\sec\frac{\pi}8\\&=&8\ln\left(\frac{2^{3/4}}{\sqrt{1+\sqrt2}}\right)\\&=&8\ln\left(2^{3/4}\right)-8\ln{\sqrt{1+\sqrt2}}\\&=&8\frac{3}{4}\ln2-8\frac{1}{2}\ln\left(1+\sqrt2\right)\\I&=&6\ln2-4\ln\left(1+\sqrt2\right)\end{eqnarray}$ .

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Tags: Integral, Math, Monday Math, Trigonometry

This entry was posted on November 15, 2010 at 12:53 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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Monday Math 143

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