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ERJAEE, G. (20)

Lecture 12

Department

MathematicsCourse Code

MATH 2BProfessor

ERJAEE, G.Lecture

12This

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MATH 2B - Lecture 12 - Integration by Parts

Every differentiation rule has a corresponding integration rule. The integration by parts

corresponds to the product rule for differentiation.

Product Rule: [𝑓(𝑥)𝑔(𝑥)]′ = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

Integral:

∫ [𝑓(𝑥)𝑔(𝑥)]′𝑑𝑥

= ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥

or 𝑓(𝑥)𝑔(𝑥) + 𝑐 = ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥

● Here we ignore the constant “c” on the left-hand side because the two indefinite

integrals on the right will give us another two constants later

● So we have: 𝑓(𝑥)𝑔(𝑥) = ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥

● We rearrange to obtain:

∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥

Now we simplify the formula by using new notations

● Let 𝑢 = 𝑓(𝑥) and 𝑣 = 𝑔(𝑥) then 𝑑𝑢 = 𝑓′(𝑥)𝑑𝑥 𝑑𝑣 = 𝑔′(𝑥)𝑑𝑥

● The formula becomes

∫ 𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫ 𝑣 𝑑𝑢

● That means instead of solving ∫ 𝑢 𝑑𝑣 directly, we can use integration by parts to

arrive at the right-hand side, and solve ∫ 𝑣 𝑑𝑢 which could be simpler

Example

∫ 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥

Solution:

𝑢 = 𝑥 𝑑𝑣 = 𝑠𝑖𝑛𝑥 𝑑𝑥

𝑑𝑢 = 𝑑𝑥 𝑣 = = 𝑐𝑜𝑠𝑥

∫ 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥 = 𝑥(−𝑐𝑜𝑠𝑥) − ∫ − 𝑐𝑜𝑠𝑥 𝑑𝑥

u dv u v v du

= −𝑥𝑐𝑜𝑠𝑥 + ∫ 𝑐𝑜𝑠𝑥 𝑑𝑥

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