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What will be the value of this integral? Please explain with proper steps!

What will be the value of this integral? Please explain with proper steps!

Question Image
Grade:12

1 Answers

Nandana
110 Points
3 years ago
hi !
  Ans : 5/4√ π erfi(sinx)-1/2 sinx esin2 x
     where erfi(sinx) is imaginary error in integration , which is ∫2/ √π eu2 du
EXP :
        ∫cosx e^sin2 x (1+cos2 x) dx
            sinx = y
            cosx dx = dy
                    =  ∫ e^y2 (2- y2 ) dy
                    =  2∫ e^y2  dy -∫ y2 e^y2  dy
               2∫ e^y2  dy    = √π ∫ 2/√π e^y2  dy
                                      = √π erfi (sinx)
            
                ∫ y2 e^y2  dy by using integration by part method
                 
                  u =y & dv = ye^y2 dy
                du = dy  v= ½  e^y2
               uv -∫v du  = ½ y e^y- ½∫ e^y2 dy
                                = ½ sinx e^sin2 x - √π/4 ∫ e^y2 dy
                                =½ sinx e^sin2 x - √π/4erfi(sinx)
 
             now, the overall integration is -
              
           
              2∫ e^y2  dy -∫ y2 e^y2  dy  = √π erfi (sinx) -[½ sinx e^sin2 x - √π/4erfi(sinx)]
                                                           
                                                          => 5√π /4 erfi(sinx) - ½ sinx e^sin2 x + K
                                                         where , K is orbitary constant
                     

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