Integration Techniques
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This page is intended to serve two purposes:
- To demonstrate the different integration techniques
- To explain when to apply each of the techniques
Contents
Substitution
Two rules of thumb for the substitution techniques:
- treat this as a default technique, ie, always try substitution first
- let u = "the most complicated term". For example:
- Q: Integrate ∫ (6 + 6x - 5)/(2 + 3 -5x - 1) dx.
- A: The easy way is to let u = 2 + 3 -5x - 1 and try it out (the hard way is to use partial fraction and try to factorize the denominator). We have du = (6 + 6x - 5) dx, and sub it in to get answer ln |(2x^{3} + 3 -5x - 1)| + C
However, there are exceptions to the rules of thumb:
- Q: Integrate ∫ x/(1 + ) dx.
- A: First substitute u = x^{2}. We know du = 2x dx, so the integral becomes ∫ 1/2(1 + u^{2}). Now use the arctan formula to get 1/2 arctan u. The answer is 1/2 arctan (x^{2}) + C.
Integration by Parts
Integration by parts come from product rule of differentiation. Recall production rule of differentiation of two functions (uv)' = u'v + uv'. We can rearrange the terms to u'v = (uv)' - uv' and integrate both sides. We get ∫ u'v = uv - ∫ uv'.
- Q: Ingegrate x**dx
- A:
First break it into two parts, one part is x*dx, the other part is .
so du=*dx, v=x
then, we can calculate u=, v=dx
so ∫ u'v=uv-∫ uv'
=xe^{x}-∫ dx
=x-
=(x-1)* + C
Partial Fraction
Partial fraction method is used when we are integrating rational functions, ie, a polynomial divided by a polynomial.
However, note that partial fraction method only works well when the degree in the numerator is smaller than the degree in the denominator. When we get the other case (degree numerator > degree denominator), we have to do long division. For example:
- Q: ∫ f(x) dx, where f(x) = (2 - 2 + 9x -1)/( - x - 2)
- A: As the numerator is of degree 3, which is larger than the degree-2 denominator, we have to do long division to get f(x) = 2x + (5x-1)/(x^{2} - x - 2). Then we want to break down the rational function (5x-1)/( - x - 2) = (5x - 1)/(x-2)(x+1) = A/(x-2) + B(x+1). Solving the equation by substituting x = 2 and x = -1 to get (5x-1)/( - x - 2) = 3/(x-2) + 2/(x+1). So the integral is ∫ 2x + 3/(x-2) + 2/(x+1) dx = + 3ln|x - 2| + 2ln|x+1| + C
Trigonometric Substitution
Trigonomitric substitution is used when we have trigonometric functions or the problem takes the form + , - , or - .
(try variations like completing square)
And there are two general types of trig integrals
A. calculate ∫ (sinx)^{m}(cosx)^{n}dx where m,n are both positive integers
A1. n is odd, m doesn't matter.
- Q: ∫ (sinx)^{2}(cosx)^{3}dx
- A:we can isolate one cos(x) from the function, and then use integration by parts.
∫ (sinx)^{2}*(cosx)^{3}dx
=∫ (sinx)^{2}(cosx)^{2}cosx dx
use (sinx)^{2}+(cosx)^{2}=1
so that (cosx)^{2}=1-
if u=sinx, du= cosx*dx
then the original function will be
=∫ *(1-)*du
Notes: The strategy is to isolate cosx, and use (cosx)^{2}=1- (sinx)^{2}
Denote u=sin(x), du=cos(x)*dx
so the function will become ∫ polynomial(u)*du
A2. m is odd, n doesn't matter
- Q: ∫ (sinx)^{5}(cosx)^{3}dx
- A: we can isolate sinx from the function and then use integration by parts.
=∫ (sinx)^{4}(cosx)^{3}sinxdx
Denote u=cosx, du=sinxdx
=-∫ (sinx)^{4}*(cosx)^{3}dx
=-∫ (1-)^{2}**du
since sinx^{2}=1-cosx^{2}=1-u^{2}
Notes: The strategy is to isolate sinx, and use (sinx)^{2}=1-(cosx)^{2}
Denote u=cosx, du=-sin(x)dx
A3. Both m and n are even numbers
Strategy: Use (cosx)^{2}=(cosx+1)/2, (sinx)^{2}=(1-cosx)/2
- Q:
∫ (sinx)^{2}*(cosx)^{2}dx
- A:
=∫ (cos2x+1)(sin2x+1)dx
=1/4 ∫ (1-cos2x)^{2}dx
=(1/4)x- 1/4 ∫ cos2x*dx
=(1/4)x-1/4 ∫ (1+cos4x)/2*dx
=(1/4)x-1/2x-1/8 ∫ cos4x*dx
=-(1/4)x-(1/8)(sin4x)*(1/4) +C
Challenging problems
- Q: Solve the following problem by both substitution and integration by parts: ∫ sin(sqrt(x)) dx.
- A: First we make a substitution u = sqrt(x), ie, du = 1/[2sqrt(x)] dx = dx/2u. Substitute this in to get int 2u*sin (u) du. Integration by part gives -2ucos(u) + int cos(u) du = -2u cos u + sin(u)+ C. Sub u = sqrt(x) back to get the answer.
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