- Integration by Recognition

The table below shows some integrals of important functions. These should be learnt since you will need to be able to INTEGRATE BY RECOGNITION.


INTEGRATION REQUIRING LOGARITHMS

LESSON 1
Determine
(i) \(\displaystyle \int {2\over x}\:dx\)
(ii) \(\displaystyle \int {4\over 4x-1}\:dx\)
(iii) \(\displaystyle \int {5\over 1-2x}\:dx\)
(iv) \(\displaystyle \int {2x+3\over 2x^2+6x-9}\:dx\)
(v) \(\int \tan ⁡x\: dx\)
SOLUTION
(i) \(\displaystyle \int {2\over x}\: dx\)
  • Rewrite integral
\(\displaystyle ⟹2\int {1\over x} dx\)
  • Integrate using logarithmic integration
\(⟹2 \ln⁡ |x|+c\)
  • Optionally, the integral can be rewritten using the properties of logarithms.
\(\displaystyle \int {2\over x} dx=\ln⁡ (x^2)+c\)
(ii) \(\displaystyle \int {4\over 4x-1} dx=\ln⁡|4x-1|+c\)
Since \(f(x)=4x-1\) and \(f'(x)=4\)

(iii) \(\displaystyle \int {5\over 1-2x} dx\)
  • We need to rewrite the integrand so that it is of the form
\[\int {f'(x)\over f(x)}\]
  • Since \(u(x)=1-2x → u'(x)=-2\) we change the numerator to \(-2\) and make the necessary adjustments.
\(\displaystyle \int {5\over 1-2x} dx=-{5\over2} \int {-2\over 1-2x} dx\)
\(\displaystyle =-{5\over2} \ln⁡|1-2x|+c\)

(iv) \(\displaystyle \int {2x+3\over 2x^2+6x-9} dx\)
  • Rewriting the integrand in the form \(\displaystyle \int {f'(x)\over f(x)}\) we get
\(\displaystyle \int {2x+3\over 2x^2+6x-9} dx={1\over2} \int {4x+6\over 2x^2+6x-9} dx\)
\(\displaystyle ={1\over 2} \ln⁡|2x^2+6x-9|+c\)

(v) \(\int \tan ⁡x\: dx\)
First of all we will rewrite \(\displaystyle \tan ⁡x\) as \(\displaystyle {\sin ⁡x\over \cos⁡ x}\) and then adjust the integrand to the necessary form.
\(\displaystyle \int \tan ⁡x \:dx=\int {\sin⁡ x\over \cos⁡x} dx\)
\(\displaystyle =-\int {-\sin ⁡x\over \cos⁡ x} dx\)
\(\displaystyle =-\ln⁡|\cos⁡ x|+c\)

INTEGRATION OF THE FORM \(\int f'(x)[f(x)]^n dx\)
LESSON 1
Determine
(i) \(\displaystyle \int 12x^3(3x^4+5)^2 \:dx\)
(ii) \(\displaystyle \int {3\over x}(2+\ln ⁡x)^3\: dx\)
SOLUTION
(i) \(\int 12x^3(3x^4+5)^2\:dx\)
  • By RECOGNITION we should realise that we have \(\displaystyle \int f'(x)[f(x)]^n\:dx=\left[{[f(x)]^{n+1}\over n+1}\right]+c\)
\(f(x)=3x^4+5 → f'(x)=12x^3\)
\(\displaystyle \int 12x^3(3x^4+5)^2\:dx={(3x^4+5)^3\over 3}+c\)
Alternately a substitution can be used.

(ii) \(\displaystyle \int {3\over x} (2+\ln ⁡x)^3 \:dx\)
\(\displaystyle f(x)=2+\ln ⁡x → f'(x)={1\over x}\)
\(\displaystyle \int {3\over x} (2+\ln ⁡x)^3\:dx=3\int {1\over x} (2+\ln ⁡x)^3\: dx\)
\(\displaystyle ={3(2+\ln ⁡x)^4\over 4}+c\)

INTEGRATION OF EXPONENTIAL FUNCTIONS
LESSON 2
Evaluate each of the following.
(i) \(\int e^{2x} dx\)
(ii) \(\int 2e^{3-x}\:dx\)
(iii) \(\int e^x(2+e^x)^3\: dx\)
(iv) \(\int e^{x+e^x}\: dx\)
(v) \(\int -\csc^2⁡x e^{\cot ⁡x}\:dx\)
SOLUTION
(i) \(\int e^{2x}\: dx\)
\(f(x)=2x → f'(x)=2\)
\(\displaystyle \int e^{2x} \: dx={1\over2} e^{2x}+c\)

(ii) \(\int 2e^{3-x}\: dx\)
\(f(x)=3-x → f'(x)=-1\)
\(\int 2e^{3-x}\:dx=2\int e^{3-x}\: dx\)
\(\displaystyle =2\left({1\over -1} e^{3-x}\right)+c\)
\(=-2e^{3-x}+c\)

(iii) \(\int e^x (2+e^x)^3\: dx\)
Though exponential integration is required the main form of integration is \(\int f'(x)[f(x)]^n\: dx\)
\(f(x)=2+e^x → f'(x)=e^x\)
\(\displaystyle \int e^x(2+e^x)^3\:dx={(2+e^x)^4\over4}+c\)

(iv) \(\int e^{x+e^x}\: dx
  • Using properties of Indices
\(\int e^x e^{e^x} dx
  • We now have \(\int f'(x) e^{f(x)}\: dx\). Therefore
\(\int e^{x+e^x}\:dx=e^{e^x}+c\)

(v) \(\int -\csc^2⁡x e^{\cot ⁡x}\:dx\)
  • We now have \(\int f'(x) e^{f(x)}\:dx\). Therefore
\(\int -\csc^2⁡x e^{\cot ⁡x}\:dx=e^{\cot ⁡x}+c\)

INTEGRALS INVOLVING INVERSE TRIGONOMETRIC FUNCTIONS
LESSON 1
Determine
(i) \(\displaystyle \int \left(x+{1\over \sqrt{4-x^2}}\right )\: dx\)
(ii) \(\displaystyle \int \left({1\over (2x-1)^3} -{3\over 9+4x^2}\right)\: dx\)
(iii) \(\displaystyle \int {1\over 1-x^2} e^{\sin^{-1}⁡x}\: dx\)
SOLUTION
(i) \(\displaystyle \int \left(x+{1\over \sqrt{4-x^2}}\right) \:dx=\int \left(x+{1\over \sqrt{2^2-x^2}}\right) \:dx\)
We have \(a=2\) and \(u=x→u'=1\), therefore comparing with the result
\(\displaystyle \int {u'\over \sqrt{a^2-u^2}}\: dx=\sin^{-1}⁡ \left({u\over a}\right)+c\)
We can determine the required integral.
\(\displaystyle \int \left(x+{1\over \sqrt{4-x^2}}\right)\:dx={x^2\over 2}+\sin^{-1}\left({x\over2}\right)+c\)

(ii) \(\displaystyle \int \left({1\over (2x-1)^3} -{3\over(3^2+(2x))^2} \right)\:dx\)
For \(\displaystyle {3\over(3^2+(2x))^2}, a=3,u=2x→u'=2\). Therefore
\(\displaystyle \int\left({1\over (2x-1)^3} -{3\over (9+4x^2)}\right) \:dx=\int {1\over (2x-1)^3}\: dx-{3\over2} \int {2\over 3^2+(2x)^2}\: dx\)
\(\displaystyle =\int (2x-1)^{-3}\: dx-{3\over2} \int {2\over 3^2+(2x)^2}\: dx\)
Using the results \(\displaystyle \int (ax+b)^n \:dx={1\over a} \left[{(ax+b)^{n+1}\over n+1}\right]+c\) and \(\displaystyle \int {u'\over a^2+u^2}\: dx={1\over a} \tan^{-1} \left({u\over a}\right)+c\)
\(\displaystyle \int \left({1\over (2x-1)^3}-{3\over (9+4x^2)}\right) dx={1\over2} \left[{(2x-1)^{-2}\over -2}\right]-{3\over2} \left[{1\over3} \tan^{-1}\left({2x\over3}\right)\right]+c\)
\(\displaystyle =-{1\over4} (2x-1)^{-2}-{1\over2} \tan^{-1}\left({2x\over3}\right)+c\)

(iii) \(\displaystyle \int {1\over \sqrt {1-x^2}} e^{\sin^{-1}⁡x}\: dx\)
We have another instance where we can use \(\int f'(x)e^{f(x)}\:dx=e^{f(x)}+c\)
\(\displaystyle f(x)=\sin^{-1}⁡x→f'(x)={1\over \sqrt {1-x^2}}\)
\(\int {1\over \sqrt{1-x^2}}e^{\sin^{-1}}⁡x\:dx=e^{\sin^{-1}⁡x} +c\)