http://www.mapleprimes.com/questions/40811-Trigonometric-Integral en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Thu, 11 Jun 2026 14:37:05 GMT Thu, 11 Jun 2026 14:37:05 GMT The latest answers and comments added to the Question, Trigonometric Integral http://www.mapleprimes.com/images/mapleprimeswhite.jpg http://www.mapleprimes.com/questions/40811-Trigonometric-Integral http://www.mapleprimes.com/questions/40811-Trigonometric-Integral?ref=Feed:MaplePrimes:Trigonometric Integral:Comments#answer75576 I used pre and /pre, but it was unable to show. Sorry. <pre> value(J); a less than -1, undefined, otherwise / / (1/2) \\ | | / 2 \ || | | |a - 1| || | ln|a + |------| a|| | | | 2 | || | \ \ a / /| | -----------------------| | (1/2) | | / 2 \ | | |a - 1| | | a |------| | | | 2 | | \ \ a / / </pre> Why a less than -1 undefined? I used pre and /pre, but it was unable to show. Sorry. <pre> value(J); a less than -1, undefined, otherwise / / (1/2) \\ | | / 2 \ || | | |a - 1| || | ln|a + |------| a|| | | | 2 | || | \ \ a / /| | -----------------------| | (1/2) | | / 2 \ | | |a - 1| | | a |------| | | | 2 | | \ \ a / / </pre> Why a less than -1 undefined? 75576 Sat, 22 Sep 2007 00:19:34 Z SandorSzabo SandorSzabo http://www.mapleprimes.com/questions/40811-Trigonometric-Integral?ref=Feed:MaplePrimes:Trigonometric Integral:Comments#answer75575 <pre> try the following to find an explanation: Int(1/((1+a*y)*(1-y^2)^(1/2)),y = 0 .. 1); subs(a=-2,%); plot( op(%)); and Int(1/((1-2*y)*(1-y^2)^(1/2)),y = 0 .. 1/2); value(%); Int(1/((1-2*y)*(1-y^2)^(1/2)),y = 1/2 .. 1); value(%); <pre> try the following to find an explanation: Int(1/((1+a*y)*(1-y^2)^(1/2)),y = 0 .. 1); subs(a=-2,%); plot( op(%)); and Int(1/((1-2*y)*(1-y^2)^(1/2)),y = 0 .. 1/2); value(%); Int(1/((1-2*y)*(1-y^2)^(1/2)),y = 1/2 .. 1); value(%); 75575 Sat, 22 Sep 2007 00:59:58 Z Axel Vogt Axel Vogt http://www.mapleprimes.com/questions/40811-Trigonometric-Integral?ref=Feed:MaplePrimes:Trigonometric Integral:Comments#answer75566 hm ... it does not depend on a = 2 or not, it is simply that the integrand has a pole for such an a (where it does not matter if you take your original problem or that with changed variables) it is (more or less) the same as integrating 1/x from some negative up to some positive value BTW: it is *really* nice how Maple answers the problem! hm ... it does not depend on a = 2 or not, it is simply that the integrand has a pole for such an a (where it does not matter if you take your original problem or that with changed variables) it is (more or less) the same as integrating 1/x from some negative up to some positive value BTW: it is *really* nice how Maple answers the problem! 75566 Sat, 22 Sep 2007 17:51:53 Z Axel Vogt Axel Vogt http://www.mapleprimes.com/questions/40811-Trigonometric-Integral?ref=Feed:MaplePrimes:Trigonometric Integral:Comments#comment84986 Thanks. Sorry for me, I forgot to say, I'm interested in principal value also, if exists. if a = -2 <pre> assume(0 less than epsi,epsi less than 0.25); Int(1/((1-2*y)*sqrt(1-y^2)),y=0..1/2-epsi)=int(1/((1-2*y)*sqrt(1-y^2)),y=0..1/2-epsi); /1/2 - epsi | 1 1 (1/2) / (1/2) \ | ----------------------- dy = - - 3 ln\2 3 + 3/ | (1/2) 6 /0 / 2\ (1 - 2 y) \1 - y / 1 (1/2) / (1/2) \ + - 3 ln\2 3 - 3/ 6 / / (1/2) \\ 1 (1/2) | | (3 + 2 epsi) 3 || + - 3 Re|arctanh|------------------------------|| 3 | | (1/2)|| | | / 2 \ || \ \3 \-4 epsi + 3 + 4 epsi/ // Int(1/((1-2*y)*sqrt(1-y^2)),y=1/2+epsi..1)=int(1/((1-2*y)*sqrt(1-y^2)),y=1/2+epsi..1); /1 / / | 1 1 (1/2) | | | ----------------------- dy = - 3 Re|arctanh| | (1/2) 3 | | /1/2 + epsi / 2\ | | (1 - 2 y) \1 - y / \ \ (1/2) \\ (-3 + 2 epsi) 3 || ------------------------------|| (1/2)|| / 2 \ || 3 \-4 epsi + 3 - 4 epsi/ // with(MultiSeries): Arctanh( 1/3*(-3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3-4/Delta))=asympt( arctanh( 1/3*(-3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3-4/Delta)), Delta,2); / / 2 \ (1/2) \ | |-3 + -----| 3 | | \ Delta/ | 1 1 /8\ Arctanh|-----------------------------| = - - ln(2) - ln(Delta) + - ln|-| | (1/2)| 2 2 \9/ | / 4 4 \ | |3 |- ------ + 3 - -----| | | | 2 Delta| | \ \ Delta / / 1 / 1 \ + - I Pi + O|-----| 2 \Delta/ Arctanh( 1/3*(3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3+4/Delta))=asympt( arctanh(1/3*(3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3+4/Delta)),Delta,2); / / 2 \ (1/2) \ | |3 + -----| 3 | | \ Delta/ | 1 1 /8\ Arctanh|-----------------------------| = - ln(2) + ln(Delta) - - ln|-| | (1/2)| 2 2 \9/ | / 4 4 \ | |3 |- ------ + 3 + -----| | | | 2 Delta| | \ \ Delta / / 1 / 1 \ - - I Pi + O|-----| 2 \Delta/ -1/6*sqrt(3)*ln(2*sqrt(3)+3)+1/6*sqrt(3)*ln(2*sqrt(3)-3)+1/3*sqrt(3)*(-1/2*ln(2)-ln(Delta)+1/2*ln(8/9)+1/2*I*Pi+1/2*ln(2)+ln(Delta)-1/2*ln(8/9)-1/2*I*Pi); 1 (1/2) / (1/2) \ 1 (1/2) / (1/2) \ - - 3 ln\2 3 + 3/ + - 3 ln\2 3 - 3/ 6 6 </pre> If a is not equal to -2 but less than -1 <pre> assume(a less than -1); Int(1/((1+a*y)*sqrt(1-y^2)),y=1/2+epsi..1)=int(1/((1+a*y)*(sqrt(1-y^2))),y=1/2+epsi..1); /1 | 1 | ----------------------- dy = - | (1/2) /1/2 + epsi / 2\ (1 + a y) \1 - y / / / / a + y \ 1 \\ -2 |limit|arctanh|---------------------------|, y = - + epsi, right|| + I Pi | | | (1/2) (1/2)| 2 || | | |/ 2 \ / 2\ | || \ \ \\a - 1/ \1 - y / / // ---------------------------------------------------------------------------- (1/2) / 2 \ 2 \a - 1/ Int(1/((1+a*y)*sqrt(1-y^2)),y=0..1/2-epsi)=int(1/((1+a*y)*sqrt(1-y^2)),y=0..1/2-epsi); /1/2 - epsi | 1 1 / | ----------------------- dy = --------------- |I Pi | (1/2) (1/2) | /0 / 2\ / 2 \ | (1 + a y) \1 - y / 2 \a - 1/ \ / / a \\ + 2 Re|arctanh|-------------|| | | (1/2)|| | |/ 2 \ || \ \\a - 1/ // / / / a + y \ 1 \\\ - 2 |limit|arctanh|---------------------------|, y = - - epsi, left||| | | | (1/2) (1/2)| 2 ||| | | |/ 2 \ / 2\ | ||| \ \ \\a - 1/ \1 - y / / /// solve( (A+1/2)/(sqrt(A^2-1)*sqrt(3/4))=-1,A); -2 Arctanh((a+1/2+1/Delta)/(sqrt(a^2-1)*sqrt(1-(1/2+1/Delta)^2)))=asympt( arctanh((a+1/2+1/Delta)/(sqrt(a^2-1)*sqrt(1-(1/2+1/Delta)^2))), Delta,2); / 1 1 \ | a + - + ----- | / | 2 Delta | | Arctanh|-------------------------------------| = arctanh| | (1/2)| | | (1/2) / 2\ | | |/ 2 \ | /1 1 \ | | \ |\a - 1/ |1 - |- + -----| | | \ \ \2 Delta/ / / (1/2) (1/2) (1/2) \ / 2 \ (1/2) (1/2) 3 4 (2 a + 1)| 2 \a - 1/ 3 4 / 1 \ -----------------------| - ----------------------------- + O|------| (1/2) | 3 (a + 2) Delta | 2| / 2 \ | \Delta / 6 \a - 1/ / </pre> Thanks. Sorry for me, I forgot to say, I'm interested in principal value also, if exists. if a = -2 <pre> assume(0 less than epsi,epsi less than 0.25); Int(1/((1-2*y)*sqrt(1-y^2)),y=0..1/2-epsi)=int(1/((1-2*y)*sqrt(1-y^2)),y=0..1/2-epsi); /1/2 - epsi | 1 1 (1/2) / (1/2) \ | ----------------------- dy = - - 3 ln\2 3 + 3/ | (1/2) 6 /0 / 2\ (1 - 2 y) \1 - y / 1 (1/2) / (1/2) \ + - 3 ln\2 3 - 3/ 6 / / (1/2) \\ 1 (1/2) | | (3 + 2 epsi) 3 || + - 3 Re|arctanh|------------------------------|| 3 | | (1/2)|| | | / 2 \ || \ \3 \-4 epsi + 3 + 4 epsi/ // Int(1/((1-2*y)*sqrt(1-y^2)),y=1/2+epsi..1)=int(1/((1-2*y)*sqrt(1-y^2)),y=1/2+epsi..1); /1 / / | 1 1 (1/2) | | | ----------------------- dy = - 3 Re|arctanh| | (1/2) 3 | | /1/2 + epsi / 2\ | | (1 - 2 y) \1 - y / \ \ (1/2) \\ (-3 + 2 epsi) 3 || ------------------------------|| (1/2)|| / 2 \ || 3 \-4 epsi + 3 - 4 epsi/ // with(MultiSeries): Arctanh( 1/3*(-3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3-4/Delta))=asympt( arctanh( 1/3*(-3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3-4/Delta)), Delta,2); / / 2 \ (1/2) \ | |-3 + -----| 3 | | \ Delta/ | 1 1 /8\ Arctanh|-----------------------------| = - - ln(2) - ln(Delta) + - ln|-| | (1/2)| 2 2 \9/ | / 4 4 \ | |3 |- ------ + 3 - -----| | | | 2 Delta| | \ \ Delta / / 1 / 1 \ + - I Pi + O|-----| 2 \Delta/ Arctanh( 1/3*(3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3+4/Delta))=asympt( arctanh(1/3*(3+2/Delta)*sqrt(3)/sqrt(-4/Delta^2+3+4/Delta)),Delta,2); / / 2 \ (1/2) \ | |3 + -----| 3 | | \ Delta/ | 1 1 /8\ Arctanh|-----------------------------| = - ln(2) + ln(Delta) - - ln|-| | (1/2)| 2 2 \9/ | / 4 4 \ | |3 |- ------ + 3 + -----| | | | 2 Delta| | \ \ Delta / / 1 / 1 \ - - I Pi + O|-----| 2 \Delta/ -1/6*sqrt(3)*ln(2*sqrt(3)+3)+1/6*sqrt(3)*ln(2*sqrt(3)-3)+1/3*sqrt(3)*(-1/2*ln(2)-ln(Delta)+1/2*ln(8/9)+1/2*I*Pi+1/2*ln(2)+ln(Delta)-1/2*ln(8/9)-1/2*I*Pi); 1 (1/2) / (1/2) \ 1 (1/2) / (1/2) \ - - 3 ln\2 3 + 3/ + - 3 ln\2 3 - 3/ 6 6 </pre> If a is not equal to -2 but less than -1 <pre> assume(a less than -1); Int(1/((1+a*y)*sqrt(1-y^2)),y=1/2+epsi..1)=int(1/((1+a*y)*(sqrt(1-y^2))),y=1/2+epsi..1); /1 | 1 | ----------------------- dy = - | (1/2) /1/2 + epsi / 2\ (1 + a y) \1 - y / / / / a + y \ 1 \\ -2 |limit|arctanh|---------------------------|, y = - + epsi, right|| + I Pi | | | (1/2) (1/2)| 2 || | | |/ 2 \ / 2\ | || \ \ \\a - 1/ \1 - y / / // ---------------------------------------------------------------------------- (1/2) / 2 \ 2 \a - 1/ Int(1/((1+a*y)*sqrt(1-y^2)),y=0..1/2-epsi)=int(1/((1+a*y)*sqrt(1-y^2)),y=0..1/2-epsi); /1/2 - epsi | 1 1 / | ----------------------- dy = --------------- |I Pi | (1/2) (1/2) | /0 / 2\ / 2 \ | (1 + a y) \1 - y / 2 \a - 1/ \ / / a \\ + 2 Re|arctanh|-------------|| | | (1/2)|| | |/ 2 \ || \ \\a - 1/ // / / / a + y \ 1 \\\ - 2 |limit|arctanh|---------------------------|, y = - - epsi, left||| | | | (1/2) (1/2)| 2 ||| | | |/ 2 \ / 2\ | ||| \ \ \\a - 1/ \1 - y / / /// solve( (A+1/2)/(sqrt(A^2-1)*sqrt(3/4))=-1,A); -2 Arctanh((a+1/2+1/Delta)/(sqrt(a^2-1)*sqrt(1-(1/2+1/Delta)^2)))=asympt( arctanh((a+1/2+1/Delta)/(sqrt(a^2-1)*sqrt(1-(1/2+1/Delta)^2))), Delta,2); / 1 1 \ | a + - + ----- | / | 2 Delta | | Arctanh|-------------------------------------| = arctanh| | (1/2)| | | (1/2) / 2\ | | |/ 2 \ | /1 1 \ | | \ |\a - 1/ |1 - |- + -----| | | \ \ \2 Delta/ / / (1/2) (1/2) (1/2) \ / 2 \ (1/2) (1/2) 3 4 (2 a + 1)| 2 \a - 1/ 3 4 / 1 \ -----------------------| - ----------------------------- + O|------| (1/2) | 3 (a + 2) Delta | 2| / 2 \ | \Delta / 6 \a - 1/ / </pre> 84986 Sat, 22 Sep 2007 13:30:39 Z SandorSzabo SandorSzabo http://www.mapleprimes.com/questions/40811-Trigonometric-Integral?ref=Feed:MaplePrimes:Trigonometric Integral:Comments#comment91926 Maple will do the Cauchy principal value if you ask it to: <pre> > int(1/(1+a*y)/sqrt(1-y^2),y=0..1,CauchyPrincipalValue) assuming a &lt; -1; </pre> <maple>1/(-1+a^2)^(1/2)*Re(arctanh(a/(-1+a^2)^(1/2)))</maple> <pre> > simplify(evalc(%)); </pre> <maple>1/2*(ln(-(-1+a^2)^(1/2)-a)-ln(-a+(-1+a^2)^(1/2))) /(-1+a^2)^(1/2)</maple> I don't know why it doesn't work for the original integral: <pre> > int(1/(1+a*cos(x)),x=0..Pi/2,CauchyPrincipalValue) assuming a &lt; -1;</pre> <maple>int(1/(1+a*cos(x)),x = 0 .. 1/2*Pi,CauchyPrincipalValue)</maple> Maple will do the Cauchy principal value if you ask it to: <pre> > int(1/(1+a*y)/sqrt(1-y^2),y=0..1,CauchyPrincipalValue) assuming a &lt; -1; </pre> <maple>1/(-1+a^2)^(1/2)*Re(arctanh(a/(-1+a^2)^(1/2)))</maple> <pre> > simplify(evalc(%)); </pre> <maple>1/2*(ln(-(-1+a^2)^(1/2)-a)-ln(-a+(-1+a^2)^(1/2))) /(-1+a^2)^(1/2)</maple> I don't know why it doesn't work for the original integral: <pre> > int(1/(1+a*cos(x)),x=0..Pi/2,CauchyPrincipalValue) assuming a &lt; -1;</pre> <maple>int(1/(1+a*cos(x)),x = 0 .. 1/2*Pi,CauchyPrincipalValue)</maple> 91926 Sun, 23 Sep 2007 08:57:13 Z Robert Israel Robert Israel