Further Integration

What would 5/∞ tend to?

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What should you should look at when deciding how to integrate...

1) Can I ln it? ie in the form f'(x) / f(x) 2) Is there any trig which I could use identities in and turn into ln|f(x)| 3) Split into partial fractions and then ln 4) Is it in a standard form for me use inverse trig if not would completing the square put it in that form 5) Substitution - could use a trig one?

How do you use the inverse integrals given for arcsine and arctan to integrate something like 4/sqrt(16-3x^2)

1) Decide if you are going to be using arcsine or arctan - if there is a sqrt (square root) and a negative x^2 you will use arcsine if there is not sqrt and a positive x^2 you will use tan 2) Here we are clearly using arcsine 3) Make your integral look like the one in the formula book so things to make sure you look out for are Numerator = 1 if it does not then take the multiple out the integral Coefficient of x^2 equals 1 if it does not take out the integral 4) For this one we need to take the for out of the numerator and remove the 3 from the coefficient of x^2 So for the numberator part do this 4 x 1/sqrt(16-3x^2) For the 3x^2 part do this 4/sqrt3 x 1/sqrt(16/3)-x^2) 5) Now sub into the arcsine/arctan formula by identifying "a" here a^2 is 16/3 so "a" is 4/sqrt3 Now put into the equation so the answer is.... 4/sqrt3 acrsine(sqrt3 x / 4)

What do you do if both limits are infinity and negative infinity?

1) Draw out the graph 2) Split the integral into two integrals where you can do and finite limit (usually zero is a good one but sometimes that can be an asymptote to use your graph sketch to check) and an infinite limit for each integral 3) Now you have two integrals each with ONE infinite limit so you can use the method you learnt for the ONE indefinite integral 4) Add integrals together

Would 100x + 25/(3x-1)(2x+1)^2 use a linear numerator in the partial fractions?

No as it is (2x-1)^2 not 2x^2 - 1 instead this would be converted to A/3x-1 + B/(2x+1)^2 + C/2x+1

Do you take the positive or negative root for arcsine when differentiating?

Positive because if you draw the graph to gradient is always positive

How do you integrate where one of the limits is an asymptote... For example if you wanted to find the integral between 1 and 3 for the graph 1/(x-2)^2

1) Draw the graph and identify the asymptotes where you are gonna find problems with the area. With this graph we can see the asymptote is at x = 2 as when x = 2 the denominator = 0 which is undefined 2) Split your integral up so you have a finite limit on one side and an infinite limit on the other so here we can do 1 to 2 where 2 is infinite and 1 is finite and then 2 to 3 again where 2 is indefinite and 3 is definite 3) For these you should have two integral which have limits of Definite 1 to indefinite 2 then Indefinite 2 to definite 3 4) Use method we used before when you have ONE indefinite integral and add together

When would you get an improper integral?

1) If one or both of your limits tend to infinity or negative infinity 2) If one or both of your limits tends to an asymptote

If you were using the inverse integrals for arcsine or arctan what form does your integral have to be in?

1) Numerator = 1 if it does not then take the multiple out the integral 2) Coefficient of x^2 equals 1 if it does not take out the integral

If you can not use arcsine or arctan to help integrate a fraction and you also can not use ln(x) what should you do?

1) Split into partial fraction Or 2) If the denominator is a quadratic then complete the square and you may be able to get it into the arcsine or arctan form

How do use integration to find the mean value for an infinite number (expression) e.g find the mean of 4/sqrt(2+3x) over the interval [2,6]

1/difference in interval x the integral of the expression using the integrals as the limits E.g 1/6-2 x ∫ 4/sqrt(2+3x) dx (limits 6 to 2)

If you have a fraction that you need to integrate with a constant on the top and a quadratic on the bottom and you cant split into partial fractions what method should you use?

Complete the square of the quadratic - if the quadratic is in a sqrt this is a huge clue to use this method This will turn it into the form to use tan or sine - remember you will get something like (x-2) + 3 at the bottom so sub in x = x-2 into the arcsine(x/a) You will need to adjust to make sure the numerator is 1 and coefficient of the x bracket bit is also 1

When do you know that you have to make the numerator a linear form ie Bx + C

If one of your denominators has an x^2 coefficient which can not be factorised For example 9x-8/(x+2)(x^2+9) would be converted to A/x+2 + Bx + C/x^2 + 9

How do you know when you can use a trig sub

If you have anything in the form a - x^2 or a + x^2

Normally when you do partial fractions what form is the numerator in once you have split the fraction?

It is a constant in the form A B or C usually e.g 3/2x+1 - 2/x-8

What would ln(∞) tend to?

It would not converge to anything so if you got this in the integral it would be divergent

What would 5^(-∞) tend to?

It would tend to zero Because... 5^(-∞) = 1/5 ∞ = 1/ large number = 0

Do you take the positive or negative root for arccosine when differentiating?

Positive root because if you draw the arccosine graph, the gradient is always negative. So you end up with -1/ positive root which is negative. If you took the negative root you would end up with -1/ negative root = positive

What do you do if ONE of the limits is infinity or negative infinity?

Turn the infinity limit into and "a" and integrate as you would normally by subbing in the top limit takeaway the bottom limit. When you are looking at what the "a" substitution is you are finding what would happen as "a" tends to infinity (or negative infinity) If you find a solution when "a" tends to infinity it means the integral is convergent If you can't find a solution where "a" tends to infinity then it i divergent

How do you decide what trigonometric substitution to make?

Use x = asin(u) if it is square root or x = atan(u) for no square root Use sine sub for a negative x^2 Use tan sub for positive x^2

How do you differentiate arccos arcsin or arctan?

Using implicit differentiation

What is an improper integral?

Where one of or both of the limits on the integral are indefinite. E.g one of the limits is infinity or on the asymptote.

What will you have to use in the trig substitution integration

You will have to use the trig indemnity sin^2 + Cos^2 = 1

Differentiate arcsine

e.g y = arcsin siny = x Implicit differentiation... cosy dy/dx = 1 so dy/dx = 1/cosy Use sin^2x + cos^2 = 1 = 1/sqrt(1-sin^2x) = 1/sqrt(1-x^2)

If you had a to use a trig substitution for sqrt(16 - x^2) what would you use?

x = 4sinu

If the denominator of the fraction was (1-x^2)^3/2 what would your substitution be?

x = sinu

If you had to use a trig substitution for (1 + x^2)^3/2 what would you use

x = tanu