INTEGRATION BY PARTS EXAMPLE PROBLEMS

Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly.

Formula :

∫udv  =  uv - ∫vdu

By differentiating "u" we get "du", by integrating dv we get v.

Note :

In order to avoid applying the integration by parts two or more times to find the solution, we may us Bernoulli’s formula to find the solution easily.

∫udv  =  uv - u'v1 + u''v2 - u'''v3 + ...............

By differentiating "u" consecutively, we get u', u'' etc. In a similar manner by integrating "v" consecutively, we get v1, v2,.....etc.

Problem 1 :

Integrate the following with respect to x:

 9xe3x

Solution :

  =  ∫ 9xe3x dx

u  =  x            dv  =  e3x

du  = dx          v  =  e3x/3

∫udv  =  uv - ∫vdu  

∫ 9 x e3x dx =  x(e3x/3) - ∫(e3x/3) dx

  =  (x/3)e3x - (1/3)e3x dx

  =  (x/3)e3x - (1/3)(e3x/3) + c

  =  (x/3)e3x - (1/9)e3x + c

  =  (e3x/3) (3x - 1) + c 

Problem 2 :

Integrate the following with respect to x:

 x sin 3x

Solution :

  =  ∫  x sin 3x  dx

u  =  x            dv  =  ∫ sin 3x

du  = dx          v  =  -cos 3x/3

∫udv  =  uv - ∫vdu  

  x sin 3x  dx  =  x(-cos 3x/3) - ∫(-cos 3x/3) dx  

  =  (-x/3) cos 3x + (1/3) ∫ cos 3x dx

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

  =  (-x/3) cos 3x + (1/9) sin 3x + c

Problem 3 :

Integrate the following with respect to x:

 25 xe-5x

Solution :

  =  ∫ 25 xe-5x  dx

u  =  x            dv  =  ∫ e-5x

du  = dx          v  =  -e-5x/5

∫udv  =  uv - ∫vdu  

 25 xe-5x  dx  =  25[x(-e-5x/5) - ∫(-e-5x/5) dx

  =  25[(-x/5)(e-5x) + (1/5) e-5x dx

  =  25[(-x/5)(e-5x) + (1/5) (-e-5x/5)] + c

  =  (25/5)(e-5x)[-x - (1/5)] + c

  =  -e-5x(5x + 1) + c

Problem 4 :

Integrate the following with respect to x:

 x sec x tan x

Solution :

  =  ∫x sec x tan x dx

u  =  x            dv  =  ∫ sec x tan x

du  = dx          v  =  sec x

∫udv  =  uv - ∫vdu  

x sec x tan x dx  =  x sec x - ∫sec x dx

=  x sec x - log (sec x + tan x) + c

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