INTEGRATION OF e POWER ax Sin bx or Cos bx

Question 1 :

Integrate the following with respect to x

 e^ax cosbx

Solution :

 ∫ e^ax cosbx dx  =  e^ax/(a² + b²) (a cos bx + b sin bx)

Question 2 :

Integrate the following with respect to x

 e^2x sin x

Solution :

 ∫ e^ax sin bx dx  =  e^ax/(a² + b²) (a sin bx - b cos bx)

 ∫  e^2x sin x dx 

a  =  2 and b = 1

  =  e^2x/(2² + 1²) (2 sin x - 1 cos x) + c

  =  (e^2x/5) (2 sin x - 1 cos x) + c

Question 3 :

Integrate the following with respect to x

 e^-x cos 2x

Solution :

 ∫ e^ax cosbx dx  =  e^ax/(a² + b²) (a cos bx + b sin bx)

 ∫  e^-x cos 2x dx 

a  =  -1 and b = 2

  =  e^-x/((-1)² + 2²) (-1 cos 2x + 2 sin 2x) + c

  =  (e^-x/5) (-cos 2x + 2 sin 2x) + c

  =  (e^-x/5) (2 sin 2x - cos 2x) + c

Question 4 :

Integrate the following with respect to x

 e^-3x sin 2x

Solution :

 ∫ e^ax sin bx dx  =  e^ax/(a² + b²) (a sin bx - b cos bx)

 ∫ e^-3x sin 2x dx 

a  =  -3 and b = 2

  =  e^-3x/((-3)² + 1²) (-3 sin x - 2 cos x) + c

  =  (e^-3x/10) (-3 sin x - 2 cos x) + c

  =  (-e^-3x/10) (3 sin x + 2 cos x) + c

Question 5 :

Integrate the following with respect to x

 e^-4x sin 2x

Solution :

 ∫ e^ax sin bx dx  =  e^ax/(a² + b²) (a sin bx - b cos bx)

 ∫ e^-4x sin 2x dx 

a  =  -4 and b = 2

  =  e^-4x/((-4)² + 1²) (-4 sin x - 2 cos x) + c

  =  (e^-4x/17) (-4 sin x - 2 cos x) + c

  =  (-e^-4x/17) (4 sin x + 2 cos x) + c

Question 6 :

Integrate the following with respect to x

 e^-3x cos x

Solution :

 ∫ e^ax cosbx dx  =  e^ax/(a² + b²) (a cos bx + b sin bx)

 ∫  e^-3x cos x dx 

a  =  -3 and b = 1

  =  e^-3x/((-3)² + 1²) (-3 cos x + 1 sin x) + c

  =  (e^-3x/10) (-3 cos x + 1 sin x) + c

  =  (e^-3x/10) (1 sin x - 3 cos x) + c

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