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# Long trig sub problem

More practice with a hairy trig sub problem. Created by Sal Khan.

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• At , should it be written sinθcosθ + sinθcosθ (as opposed to sinθcosθ + sinθ + cosθ)? I'm pretty sure it is..
• yes youre right! I seen it and had to replay just to make sure but he said it write just wrote it wrong .. x
• correct me if im wrong but didnt he get the problem wrong because he stopped carrying over the +C?
• No-It's kind of understood to be there (during the problem), but your final answer HAS to have the constant of integration, or C.
• In which video did he prove his integral at , my integral table shows x/2 + (sin2ax)/4a + C.
• I didn't see it proved in the trig playlist, but I just proved it and I feel so proud :)

First, Cos(2x) = Cos^2(x) - Sin^2(x) (which is proved in the trig playlist, so I don't need to prove it here). Then to get Cos^2(x) by itself you add Sin^2(x) to both sides

Cos(2x) + Sin^2(x) = Cos^2(x)
We wanna get rid of that Sin and we know Sin^2(x)= 1-Cos^2(x) (which again is already proved), so do a little replacement there and you get

Cos^2(x)=Cos(2x)+1-Cos^2(x), now this seems a little familiar, what you do next is add Cos^2(x) to both sides and now you have

(2)Cos^2(x)=1+Cos(2x), divide both sides by two and BAM, you now have

Cos^2(x)=1/2(1+cos(2x)) (I know this is a little late, but hopefully at least one other person who's unsure can look at it)

All of the trig Identities can be proved using what Sal gave us in the Trig playlist, Play around a little and you can find out some useful stuff.
• So, Sal shows sample problems for the forms (a^2 - x^2) as well as (a^2 +x^2); however, I did not see any problems with the form (x^2 - a^2). One of the practice problems I was given took this form, and I thought that hyperbolic trig substitution would be appropriate since we can use the identity cosh^2(theta) - 1 = sinh^2(theta). I arrived at a reduced answer to the problem in terms of inverse hyperbolic trig functions; however all of the multiple choice answers were in terms circular trig functions. Is there an easy substitution I'm missing?
• Here is another solution, but with an example (it might even be a question from Khan).
This is my very first LaTeX document - so, 'scuse any weirdness.
http://bajasound.com/khan/khan1.pdf
• Is integration just a process of trial and error and remembering types or are there any things to look out for and do to bring a expression into a expression that can be easily integrated ?
As there are no formulas like quotient rule or chain rule for it as in differentiation if a complicated expression to integrate is given how should it be approached ?
• There are tips and tricks for integration. For example, the power rule is (I think) the simplest integration rule. It is really the reverse of the power rule for derivatives: d/dx (x^n) = nx^(n-1)
The power rule for integrals says: ∫ x^n dx = ( x^(n+1) ) / (n+1)
There are also methods of integration like trig sub, u sub, integration by parts, partial fraction decomp...
Knowing what methods to use when just requires a lot of practice. You can probably find practice problems if you search google (Khan Academy does not have integration practice modules).
• The final result should be what you had +C. I don't know if you left that constant out on purpose, but in order for it to be in the correct form of an indefinite integral, the constant needs to be there, right?

Thanks so much, this was extremely helpful!
• There is +C, Sal just forgot to write it. He was to happy to finish this question :>
• I got the same solution as the video from doing it manually, but how come wolfram's integration calculator got (1/2)(x-3)(sqrt(-x^2+6x-5)-2arcsin((3-x)/2) instead?
• There are different ways of solving these problems. You can get very different-looking, but mathematically equivalent answers depending on how you solved the problem. This is especially the case when you have trigonometric functions involved.
• Some practice exercises will be added for u-substituiton, trig substituiton and other integrals? Thanks.
• There is always more videos than exercises, because they are produced much faster (as Sal said in one of his videos). But I am sure your patience will pay off, its just a matter of time before we get exercises of all kinds :D
• Look at the original problem. ∫√(6x-x²-5)dx Any value inside the square root that's less than 0 gives you a problem already. So 6x-x²-5 ≥ 0 already meaning 1 ≤ x ≤ 5. So replacing it with sin doesn't actually change anything.