HW SIX

1. (6.5)

   (a)

   $$I = \int\limits_0^{\Pi/2} \frac{\sin^2(x)}{(\sin(x)+\cos(x))} d x$$

   Let us calculate the value of this intergral numerically using the adaptive quadrature with machine precision. The result is

   $$I\simeq 0.6232252401402316$$

   Almost same result is obtained by evaluating analytically

   $$J = \frac{1}{2\sqrt{2}}\ln(3+2\sqrt{2})\simeq 0.62322524014023051339$$

   We therefore conclude that it is likely that the statement that $I=J$ is true.

   Note that this calculation does not provide a proof that this is true, only a demonstration that this is plausible assumption. Differnce in the last few digits are to be expected due to the complexity of this integral.

   (b)

   Similarly, ccccx

   $$I = \int\limits_0^{\Pi/2} \frac{1}{(\cos^2(x)+3\sin^2(x))^3} d x \simeq 0.453449841058554466,$$

   while

   $$J =\frac{\pi\sqrt{3}}{12}\simeq 0.45345,$$

   so we conclude that $I\simeq j$, so that this statement is also likely to be true.

   (c)

   $$I=\int\limits_0^1\frac{\tan^{-1}(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}} d x \simeq 0.514042,$$
   while
   $$\frac{5 \pi^5}{96}\simeq 15.9385,$$
   so we confidently conclude that $I\ne J$ and this statement is false.

   (d)

   $$I = \int\limits_0^{\pi/2} \cos^{-1}(\frac{\cos(x)}{1+ 2 \cos(x)}d x \simeq 2.0561675835602830461,$$
   which is close to
   $$J = \frac{5\pi^2}{24}.$$
   So this statement is again likely to be true.

   (e)

   $$I=\int\limits_0^1 x^{x^3}\simeq 0.940318,$$
   while
   $$J = 1+ \sum\limits_{k=0}^\infty\frac{(-1)^{k+1}}{(4+3 k)^{k+2}}\simeq 1+ \sum\limits_{k=0}^{1000}\frac{(-1)^{k+1}}{(4+3 k)^{k+2}}\simeq 1.05968,$$
   so we confidently conclude that $I\ne J$.

   We emphasize that using numerical methods to analyze analytic formulas of this kind can confidently proove that $I\ne J$, but can only give indications that $I=J$, with out prooving it.

2. (6.12) To evaluate the area under the curve
   $$x^4+2 y^4 =1,$$
   we need to evaluate numerically the integral
   $$I = \int\limits_{-1}^1 \Big(\frac{1-x^4}{2}\Big)^{\frac{1}{4}} d x.$$
   After using adaptive quadrature we otain the result
   $$I\simeq 1.5590847497554112301.$$