Maharashtra State BoardHSC Arts 12th Board Exam
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Methods of Integration: Integration by Parts

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notes

If u and v are any two differentiable functions of a single variable x . Then, by the product rule of differentiation, we have 
`d/(dx)(uv)` = u`(dv)/(dx)` + v`(du)/(dx)`
Integrating both sides, we get

uv = `int u (dv)/(dx) dx + int v (du)/(dx) dx`

or `int u (dv)/(dx)dx = uv - int v (du)/(dx) dx`        ...(1)

Let u = f(x) and `(dv)/(dx)`=g(x) . Then 

`(du)/(dx) = f'(x) and v = int g(x) dx`

Therefore, expression (1) can be rewritten as
`int f(x) g(x) dx = f(x)int g(x) dx - int [ int g(x) dx] f'(x) dx`

i.e. `int f(x) g(x) dx = f(x) int g(x) dx - int [f'(x) int g(x) dx] dx`

If we take f as the first function and g as the second function, then this formula may be stated as follows: 

“The integral of the product of two functions 
= (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function) × (integral of the second function)]”

description

  • `int(u.v) dx = u intv dx - int((du)/(dx)).(intvdx) dx`
  • Integral of the type ∫ ex [ f(x) + f'(x)] dx = exf(x) + C
  • Integrals of some more types
  1. `I = int sqrt(x^2 - a^2) dx = x/2 sqrt(x^2 - a^2) - a^2/2 log | x + sqrt(x^2 - a^2 )| + C`
  2. `I = int sqrt(x^2 - a^2) dx = x/2 sqrt(x^2 - a^2) - a^2/2 log | x + sqrt(x^2 - a^2 )| + C`
  3. `I = int sqrt(x^2 - a^2) dx = x/2 sqrt(x^2 - a^2) - a^2/2 log | x + sqrt(x^2 - a^2 )| + C`
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Shaalaa.com | Integrals part 30 (Integration by parts)

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Integrals part 30 (Integration by parts) [00:13:59]
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