1. (6.5)


    $\displaystyle 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

    $\displaystyle I\simeq 0.6232252401402316$

    Almost same result is obtained by evaluating analytically

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

    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.


    Similarly, ccccx

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


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

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


    $\displaystyle I=\int\limits_0^1\frac{\tan^{-1}(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}} d x
\simeq 0.514042,$


    $\displaystyle \frac{5 \pi^5}{96}\simeq 15.9385,$

    so we confidently conclude that $I\ne J$ and this statement is false.


    $\displaystyle I = \int\limits_0^{\pi/2} \cos^{-1}(\frac{\cos(x)}{1+ 2 \cos(x)}d x

    which is close to

    $\displaystyle J = \frac{5\pi^2}{24}.$

    So this statement is again likely to be true.


    $\displaystyle I=\int\limits_0^1 x^{x^3}\simeq 0.940318,$


    $\displaystyle J = 1+ \sum\limits_{k=0}^\infty\frac{(-1)^{k+1}}{(4+3 k)^{k+2}}\simeq
1+ \sum\limits_{k=0}^{1000}\frac{(-1)^{k+1}}{(4+3 k)^{k+2}}\simeq 1.05968,$

    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

    $\displaystyle x^4+2 y^4 =1,$

    we need to evaluate numerically the integral

    $\displaystyle I = \int\limits_{-1}^1 \Big(\frac{1-x^4}{2}\Big)^{\frac{1}{4}} d x.$

    After using adaptive quadrature we otain the result

    $\displaystyle I\simeq 1.5590847497554112301.$