INTEGRATION BY PARTS WITH TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS

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.

Question 1 :

Integrate the following with respect to x:

 x log x

Solution :

  =  ∫ x log x dx

u = log x            ∫dv  =  ∫ x dx

du  = 1/x          v  =  x²/2

∫udv  =  uv - ∫vdu  

∫ x log x dx  =  log x(x²/2) - ∫(x²/2) (1/x) dx

=  (x²/2)log x - (1/2) ∫x dx

=  (x²/2)log x - (1/2) (x²/2) + c

=  (x²/4)(2log x - 1) + c

Question 2 :

Integrate the following with respect to x:

27x² e^3x

Solution :

  =  ∫ 27x² e^3x dx

We have power for x term, to get the answer we may have to apply the formula for integration by parts two or three times. In order to avoid this, let us apply Bernoulli’s formula.

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

u = x²            ∫dv  =  ∫ e^3x dx

u'  =  2x          v  =  e^3x/3

u''  =  2          v₁  =  e^3x/9

                      v₂  =  e^3x/27

∫ 27x² e^3x dx  =  27[ x²(e^3x/3) - 2x(e^3x/9) + 2(e^3x/27)] 

  =  27e^3x [(x²/3) - (2x/9) + (2/27)] + c

  =  e^3x [9x² - 6x + 2] + c

Question 3 :

Integrate the following with respect to x:

x² cos x

Solution :

  =  ∫ x² cos x dx

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

u = x²            ∫dv  =  ∫ cos x dx

u'  =  2x          v  =  sin x

u''  =  2          v₁  =  -cos x

                      v₂  =  -sin x

∫ x² cos x dx  =  x²(sin x) - 2x(-cos x) + 2(-sin x)

  =  x²sin x + 2x cos x - 2 sin x + c

Question 4 :

Integrate the following with respect to x:

x³ sin x

Solution :

  =  ∫ x³ sin x dx

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

u = x³            ∫dv  =  ∫ cos x dx

u'  =  3x²          v  =  sin x

u''  =  6x          v₁  =  -cos x

u'''  =  6          v₂  =  -sin x

                      v₃  =  cos x

∫x³ sin x dx 

=  x³(sin x) - 3x²(-cos x) + 6x(-sin x) - 6(cos x)

=  x³ sin x + 3 x² cos x - 6 x sin x - 6 cos x + c

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