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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