Calculeaza Integrala

[Eusebiu1]

O integrala simpla.

Sa vedem cine reuseste primul!

[Whai_Nooa]

Integral Calculator: Wolfram Mathematica Online Integrator

[Silviu]

Eu.

[youjnhuhn2]

csc(1) log(sin(1+x))-csc(1) log(sin(2+x))+C

[Eusebiu1]

Si care e rationamentul ?

[Whai_Nooa]

Si care e rationamentul ?

Iti faci tema la mate?

[Eusebiu1]

Nu:)).Am rezolvat-o.Vreau sa vedem daca reuseste careva

[Whai_Nooa]

Nu:)).Am rezolvat-o.Vreau sa vedem daca reuseste careva

Ti-o rezolvam eu daca le invatam la liceu..but fuck..

[youjnhuhn2]

Take the integral:

integral csc(x+1) csc(x+2) dx

Write csc(x+1) csc(x+2) as 2/(cos(1)-cos(2 x+3)):

= integral 2/(cos(1)-cos(2 x+3)) dx

Factor out constants:

= 2 integral 1/(cos(1)-cos(2 x+3)) dx

For the integrand 1/(cos(1)-cos(2 x+3)), substitute u = 2 x+3 and du = 2 dx:

= integral 1/(cos(1)-cos(u)) du

For 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))) ds

Simplify 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)) ds

Factor out constants:

= 2 integral 1/(s^2+s^2 cos(1)-1+cos(1)) ds

Factor cos(1)-1 from the denominator:

= 2 integral 1/((cos(1)-1) ((s^2 (1+cos(1)))/(cos(1)-1)+1)) ds

Factor out constants:

= 2/(cos(1)-1) integral 1/((s^2 (1+cos(1)))/(cos(1)-1)+1) ds

For 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) dp

The integral of 1/(1-p^2) is tanh^(-1)(p):

= (2 tan(1/2) tanh^(-1)(p))/(cos(1)-1)+constant

Substitute back for p = s cot(1/2):

= csc(1/2) sec(1/2) (-tanh^(-1)(s cot(1/2)))+constant

Substitute back for s = tan(u/2):

= csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(u/2)))+constant

Substitute back for u = 2 x+3:

= csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(x+3/2)))+constant

Which is equivalent for restricted x values to:

Answer: |

| = csc(1) (log(sin(x+1))-log(sin(x+2)))+constant

[Eusebiu1]

Nu mai folositi Wolfram ca nu se pune.

[bodostyle]

csc(1) log(sin(1+x))-csc(1) log(sin(2+x))+C

Scuze de offtopic, dar ce-i cu saracia aia de semnatura ?

[Fame]

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