MES 2-3rd week - Integrands often arise as products of other functions.
- Integration by parts is a technique which uses the product rule in reverse
- Recall the formula of differentiation of a product:
- d(uv) = udv+ vdu.
- Integrating the both sides we have:
- dx= dx + dx, then
- dx=uv- dx - this is the formula known as integration by parts.
- The aim is that the integral on the RHS is simpler than the one on the LHS.
Integration by parts – how to apply - To apply the method of integration by parts:
- 1. Identify u(x) and v′(x) to ensure that dx is simpler than dx
- 2. Determine u′(x) and v(x) =dx
- 3. Plug these into the formula: dx=uv- dx
- 4. Do remaining integral, remembering the constant of integration when it is indefinite.
Integration by parts –example1.41 F dx Solution: we choose =u; =, then = ; =1 = -+= =-++c Pay attention that if we choose sinx=u we will get in LHS more complicated integral than the original one: Integration by parts –example1.42 dx Solution: we choose =u; =, then =; =1 = = -= +c Verify by differentiation: (+c)= +0= ,so the solution is correct dx Solution: we choose =;, then =; = dx== - Now we apply the integration by parts to again: we choose =;, then =; = -=- And finally: dx=-+ +C Integration by parts and substitution–example dx Solution: we choose =;=, then =; = dx==- Now use substitution: =y; then =-=; dx= + C Integration using Partial Fractions - Knowing how to decompose fractions will be beneficial to us when trying to integrate certain rational functions which look like
- General Strategy for Partial Fractions Decomposition:
- Divide if Improper: If is an improper fraction (if the degree of N(x) is greater than or equal to the degree of D(x)), divide D(x) into N(x) to find = polynomial +
- Factor Denominator: Completely factor the denominator into factors of the form (px + q)m and (ax2 + bx + c)n , where ax2 + bx + c is irreducible
Integration using Partial Fractions- continuation - Aa
- Dd
- Linear Factors: For each factor of the form (px + q)m, the partial fractions decomposition consists of the sum of m terms + + ... +
- Quadratic Factors: For each factor of the form (ax2 + bx + c)n , the partial fractions decomposition consists of the sum of n terms + + ... +
- Find
- Solution:
- Split=;Find A &B
- Multiply both sides by
- 3=A(x+3) + Bx;
- Set x=0 gives: 3A=3; A=1;
- Set x=-3 gives: -3B=3;B=-1
- Integrate
- = -
- Doing substitution u=x+3 for the second integral we have:
- lnx-ln(x+3) +C=ln+C
Integration using Partial Fractions- example-1 - Find
- Solution:
- At first divide ( for details see the auxiliary slide No13)
- The division gives: 4x+8+
- Now factor the denominator:⇒4x+8+;
- then as in the previous example split=; Find A &B
Integration using Partial Fractions- example-1- continuation - A(x-2)+Bx=13x+5
- Set x=2 gives: 2B=26x+5=31; Set x=0 gives:
-2A=5; hence B= : A=- - Integrate
- =- + +
- Doing substitution u=x-2 for the last integral we have:
- =22+8-lnx+ ln(x-2)+C
Examples-1 auxiliary - long division of polynomials - Divide
- For the moment, ignore the other terms and look just at the leading of the divisor and the leading 4x3 of the dividend.
- 4x
- Take that 4x, and multiply it through the divisor,
- 4x
- 3-8x2)
- 8x2
- Now we should look at the from the divisor and the new leading term, the 8x2, in the bottom line of the division.
- 4x+8 ⇙Take that 8, and multiply it through the divisor,
- 3-8x2)
- 8x2
- -(8x2)
Integration using Partial Fractions- example-2 - Example: Find
- Solution:
- At first factor the denominator:(x2 =
- =((x-2)(x+1))2= (x-2) 2(x+1)2- There are two repeated linear factors!
- Then split=++; We should find: A; B; C and D
- =A(x-2)(x+1)2 C(
Integration using Partial Fractions- example-1st continuation - Set x=-1 gives: D(-1-2) =-1-2-1⇒9D=-4; D=-
- Set x=2 gives: B(2+1)=8+4-1 ⇒9B=11; B=
- Set x=0 gives:-2A+B+4C+4D ⇒-2A++4C- ⇒-2A+4C= ⇒-A+2C= ⇒A-2C=
- Set x=1gives:-4A+4B+2C+D=1+2-1⇒2 ⇒
-4A++2C- =2 ⇒-4A+2C=- ⇒-2A+C=- ⇒ 2A-C= Integration using Partial Fractions- example-2 2nd continuation Then:⇒ ⇒3C=; C= 2A-= ⇒2A= ; A= - Integrate:
- =++
+ - =ln-+ln+ =ln- -+ln+F
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