I think I found another bug in `MapleV.4 (00f, unix version)`

a:= int(x/(sin(x)^3+cos(x)^3),x=0..Pi/2);

Maple is apparently able to do a symbolically using dilogaritms.

But,

evalf(a)= -.16989 -1.3791 I

which is wrong, being the integrand strictly positive.

Moreover, a direct numerical computation gives

a:= 1.4751.....

In Maple V Release 5 the ﬁrst calculation results in `.3784842693 + 2.758230170 I`

, while
an `evalf( Int( ... ) );`

leads to `a := 1.4751....`

It is corrected with Maple 6. (U. Klein)

The antiderivative found by MAPLE has some singularities; one is in the interval of
integration. The rest of the worksheet attached below seems to indicate that the
antiderivative is correct in `[0,Pi/4)`

.

> A := int( x/(sin(x)^3+cos(x)^3),x ): > readlib(discont): evalf( discont(A, x ) ); -9 {-2.356194490 + .2638180391 10 I, 2.356194490 - .6584789486 I, -.7853981634 + .6584789479 I, .7853981634 + .6584789479 I, -9 2.356194490 + .6584789479 I, -.7853981634 + .2638180391 10 I, -9 .7853981634 - .6584789486 I, .7853981634 + .2638180391 10 I, -.7853981634 - .6584789486 I, -2.356194490 + .6584789479 I, -9 -2.356194490 - .6584789486 I, 2.356194490 + .2638180391 10 I} > B := diff(A,x): C := abs(B - x/(sin(x)^3+cos(x)^3) ): > for i from 1 to 100 do > if > evalf(subs(x=i*Pi/400,C)) > 10.^(-7) > then print( [ i*Pi/400, evalf(subs(x=i*Pi/400,C))]) > fi > od; 49 -6 [--- Pi, .6861817978 10 ] 200 99 -5 [--- Pi, .1288606775 10 ] 400 10 [1/4 Pi, .1447924238 10 ]

Note the positive exponent of the last evaluation.

> subs(x=97*Pi/400,A) - subs(x=0,A): evalf(% ); -8 .3687071756 + .22 10 I > (evalf@Int)( x/(sin(x)^3+cos(x)^3),x = 0 .. 97*Pi/400 ); .3687071760

I think this is yet another example of trouble one can get with some integrals calculated by Maple V. As already noticed by other MUGers, the pedestrian procedure seems to be the following:

- use Int(...) instead of int(...), and THEN use evalf. Here, it gives 1.475106979 - if more than the numerical value is needed, and when Maple V suggests

an obviously wrong result, all hope is not lost. A solution DOES exists, that can be obtained (1) with paper and pencil (2) by tweaking the integral.

In the above case, one can play with the integrand and replace the x factor in the numerator
by `Pi/4`

, because of the symmetry of the denominator. Then, one can express the
denominator with respect to `t=x-Pi/4`

. Surprise (`?`

), the expression simpliﬁes greatly
as

Pi/sqrt(2)*int(1/(1-u^2)/(1+2*u^2),u=0..sqrt(1/2))

Maple V 3.0 (and 4.00b, U. Klein) for Windows can take it from there and gives the correct result, even though in a not so simple form (in my opinion). Anyway, the result is:

> evalf(Pi^2/12+Pi*sqrt(2)/6*ln(sqrt(2)+1)); 1.475106979