# Integration Techniques

This article is part of the MathHelp Tutoring Wiki |

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.

- Back to Integral Calculus
- Back to MathHelp