# 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) 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 e3x

Solution :

=  ∫ 27x2 e3x 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  =  ∫ e3x dx

u'  =  2x          v  =  e3x/3

u''  =  2          v1  =  e3x/9

v2  =  e3x/27

∫ 27x2 e3x dx  =  27[ x2(e3x/3) - 2x(e3x/9) + 2(e3x/27)]

=  27e3x [(x2/3) - (2x/9) + (2/27)] + c

=  e3x [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)

=  xsin x + 3 xcos x - 6 x sin x - 6 cos x + c Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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