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### Course: Class 12 math (India)>Unit 10

Lesson 9: Challenging definite integrals

# Challenging definite integration

A rather difficult integration by parts problem. Created by Sal Khan.

## Want to join the conversation?

• But Zero isn't an even number. From Wikipedia "An even number is an integer that is evenly divisible by 2." Shouldn't the function have had a third definition for when the function equals zero?
• 0/2 = 0. Therefore, 0 is evenly divisible by 2 and is even.
• I don't understand how the antiderivative of cos πx is 1/π sin πx. I know the antiderivative of cos x is sin x, but where does the 1/π come in? I would be grateful if anybody could point me to a resource that would help with this.
• This is best to understand by u substitution.
let u = πx
thus du = π dx
thus dx = du/π
So, ∫ cos (πx) dx = ∫ cos (u) du/π = 1/π ∫ cos (u) du
= (1/π)(sin u) + C
back substituting
= (1/π)(sin πx) + C
• I'm thoroughly confused. At , Sal gives a formula for integration by parts which he says he's gone through many times, but I just watched the integration by parts videos and didn't find it written it in this form? And I really thought I was getting the hang of these du, dv notations :(

Can someone please explain how ∫f(x)g'(x)dx=∫udv? I mean, I get it's some sort of substitution, but can't understand how f, g, and dx can be reduced to u and dv.

Any help is greatly appreciated! :)
• Nvm, I was able to derive it:
(uv)'=uv'+u'v
d(uv)/dx=u*dv/dx+du/dx*v
u*dv/dx=d(uv)/dx-du/dx*v
∫(u*dv/dx)dx=∫d(uv)/dx*dx-∫(du/dx*v)dx
∫u*dv=∫d(uv)-∫du*v
∫udv=uv-∫vdu
• why there is no "c"in the progress?even in the end, thanks in advance
• You do not use the + C with definite integrals, just with indefinite integrals. If you use the Fundamental Theorem of Calculus to compute the definite integral, there is a + C but it cancels out and we can ignore it.
For example, let us use a simple integral:
F(x)= x² + 3x + C ← the solution to an indefinite integralNow let compute this as a definite integral with bounds [2, 3]:F(3) - F(2) = [3²+(3)(3)+C] - [2²+3(2)+C]= 18 + C - 10 - C ← Notice I am adding and subtracting C.= 8

So, though you technically do have a +C when computing a definite integral via the Fundamental Theorem of Calculus, the + C always is both added and subtracted, thus dropping out. Therefore, we usually just leave out mentioning it altogether.
• The pop-up comment said the second term towards the end should be positive, not negative, but I don't see why it'd be positive... Can someone explain? I think I'm missing a step.
• Since it is minus that integral, the signs switch. The integral had a negative, so minus the integral is supposed to have a positive.
• At isn't it (x-1) not (1-x)? That would completely change the answer wouldn't it?
• It is 1-x. The slope of that line is negative.
• it will take atleast 6 to 7 min to solve it in exam but we are given only 3 min per question .is there any easier way to do it.Or should I leave these kind of question
• Practice a lot, so you know many different techniques and can tell in advance whether they will work.
Second, memorize as many integral forms as you can. After all, if you know the solution to a challenging integral based on an integral form, you don't have to figure out how to get the answer.
Here is a table of many integral forms. Just study these, maybe make flashcards, and hopefully you will be in good shape:
• At why does Sal put a negative in front of x/pi*sin(pi*x)?
• This is a mistake.
The integral of:
integral(x*cos(pi*x)dx) = xsin(pi*x)/pi + cos(pi*x)/pi^2
Source: http://www.wolframalpha.com/input/?i=integral+xcos%28pi*x%29+dx
• What is the greatest integer function? Is it the same as the absolute value?