Eusebiu1 Posted December 25, 2013 Report Share Posted December 25, 2013 O integrala simpla.Sa vedem cine reuseste primul! Quote Link to comment Share on other sites More sharing options...

Whai_Nooa Posted December 25, 2013 Report Share Posted December 25, 2013 Integral Calculator: Wolfram Mathematica Online Integrator Quote Link to comment Share on other sites More sharing options...

Silviu Posted December 25, 2013 Report Share Posted December 25, 2013 Eu. Quote Link to comment Share on other sites More sharing options...

youjnhuhn2 Posted December 25, 2013 Report Share Posted December 25, 2013 csc(1) log(sin(1+x))-csc(1) log(sin(2+x))+C Quote Link to comment Share on other sites More sharing options...

Eusebiu1 Posted December 25, 2013 Author Report Share Posted December 25, 2013 Si care e rationamentul ? Quote Link to comment Share on other sites More sharing options...

Whai_Nooa Posted December 25, 2013 Report Share Posted December 25, 2013 Si care e rationamentul ?Iti faci tema la mate? Quote Link to comment Share on other sites More sharing options...

Eusebiu1 Posted December 25, 2013 Author Report Share Posted December 25, 2013 Nu:)).Am rezolvat-o.Vreau sa vedem daca reuseste careva Quote Link to comment Share on other sites More sharing options...

Whai_Nooa Posted December 25, 2013 Report Share Posted December 25, 2013 Nu:)).Am rezolvat-o.Vreau sa vedem daca reuseste careva Ti-o rezolvam eu daca le invatam la liceu..but fuck.. Quote Link to comment Share on other sites More sharing options...

youjnhuhn2 Posted December 25, 2013 Report Share Posted December 25, 2013 Take the integral: integral csc(x+1) csc(x+2) dxWrite csc(x+1) csc(x+2) as 2/(cos(1)-cos(2 x+3)): = integral 2/(cos(1)-cos(2 x+3)) dxFactor out constants: = 2 integral 1/(cos(1)-cos(2 x+3)) dxFor the integrand 1/(cos(1)-cos(2 x+3)), substitute u = 2 x+3 and du = 2 dx: = integral 1/(cos(1)-cos(u)) duFor the integrand 1/(cos(1)-cos(u)), substitute s = tan(u/2) and ds = 1/2 sec^2(u/2) du. Then transform the integrand using the substitutions sin(u) = (2 s)/(s^2+1), cos(u) = (1-s^2)/(s^2+1) and du = (2 ds)/(s^2+1): = integral 2/((s^2+1) (cos(1)-(1-s^2)/(s^2+1))) dsSimplify the integrand 2/((s^2+1) (cos(1)-(1-s^2)/(s^2+1))) to get 2/(s^2+s^2 cos(1)-1+cos(1)): = integral 2/(s^2+s^2 cos(1)-1+cos(1)) dsFactor out constants: = 2 integral 1/(s^2+s^2 cos(1)-1+cos(1)) dsFactor cos(1)-1 from the denominator: = 2 integral 1/((cos(1)-1) ((s^2 (1+cos(1)))/(cos(1)-1)+1)) dsFactor out constants: = 2/(cos(1)-1) integral 1/((s^2 (1+cos(1)))/(cos(1)-1)+1) dsFor the integrand 1/((s^2 (1+cos(1)))/(cos(1)-1)+1), substitute p = s cot(1/2) and dp = cot(1/2) ds: = (2 tan(1/2))/(cos(1)-1) integral 1/(1-p^2) dpThe integral of 1/(1-p^2) is tanh^(-1)(p): = (2 tan(1/2) tanh^(-1)(p))/(cos(1)-1)+constantSubstitute back for p = s cot(1/2): = csc(1/2) sec(1/2) (-tanh^(-1)(s cot(1/2)))+constantSubstitute back for s = tan(u/2): = csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(u/2)))+constantSubstitute back for u = 2 x+3: = csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(x+3/2)))+constantWhich is equivalent for restricted x values to:Answer: | | = csc(1) (log(sin(x+1))-log(sin(x+2)))+constant 1 Quote Link to comment Share on other sites More sharing options...

Eusebiu1 Posted December 25, 2013 Author Report Share Posted December 25, 2013 Nu mai folositi Wolfram ca nu se pune. Quote Link to comment Share on other sites More sharing options...

bodostyle Posted December 26, 2013 Report Share Posted December 26, 2013 csc(1) log(sin(1+x))-csc(1) log(sin(2+x))+CScuze de offtopic, dar ce-i cu saracia aia de semnatura ? Quote Link to comment Share on other sites More sharing options...

Fame Posted December 26, 2013 Report Share Posted December 26, 2013 (edited) Numitorul este produs, deci provine dintr-o adunare de 2 fractii => ne gandim la [ln(ceva)]' * (ceva)' (cazul fericit )Calculand cos(x+1)/sin(x+1) - cos(x+2)/sin(x+2) face, prin aducere la numitor comun, sin1/sin(x+1)sin(x+2). Revenid la integrala initiala, aceasta devine 1/sin1*integrala[cos(x+1)/sin(x+1) - cos(x+2)/sin(x+2)]dx = 1/sin1 * {ln[sin(x+1)] - ln[sin(x+2)]}+ C.Sper ca nu am greseli, am facut-o in graba. Edited December 26, 2013 by Fame Quote Link to comment Share on other sites More sharing options...