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Integration probs. (1 Viewer)

peenuts

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I am having trouble with integrating questions, if there's anyone who can please respond with a step by step answer, that would be terrific.

Integrate:
1. Integral sinxcos^2x dx


2. (HSC 1997, Q1.e ) Integral sec3xtan3x dx

help :confused:
 

Minai

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yes they both are Extension 1 questions
and yes, the standard integrals sheet is ur best friend for that
 

Lazarus

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They are 2 unit questions and they are meant to be solved by observation alone.

Q1
It asks you to integrate sin(x).cos(x) dx. Think about what you would differentiate to get that. Work backwards, and start with the highest power - i.e. cos(x). It's difficult to explain, but you should be able to figure out then that the answer is (1/3)cos(x) + c.

Remember that when you differentiate something like cos(x), it is the same as differentiating (cos(x)). Using the chain rule for functions raised to a power, taking the power down the front and multiplying by the derivative of the function inside the brackets, you'll find that the actual derivative is 3.cos(x).sin(x). The function given in the question doesn't have a 3 out the front, so it must have been cancelled out, which is where the 1/3 comes from. The c is of course the ubiquitous constant of indefinite integration.

Q2
Again, use observation. As MinAi mentioned, the table of standard integrals might help.

Notice that the differential of (1/a)sec(ax) is sec(ax).tan(ax). The function given in the question is sec(3x).tan(3x). Working backwards, it's easy to see that the integral will be (1/3)sec(3x) + c.
 

Minai

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oops, that Q2 looked 3U to me...! :(

if they are meant to be solved by observation, they b worth just 1 or 2 marks right?
 

Big Willy

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Intergration

Thoes questions (the level of calibure) dont get asked anymore.

I did ones like cos^6xsinx (cos squared six x sin x) and harder ones with maths tutor.
Absoulute pain in the ass; and i dont think i can remember how to do it :(
But lazurus was right, treat them as (cosx)^3 and so on.

And u have to remember. un' = (n-1).(u)^n-1.u' (chain rule) (use this to work backwards to verify that ur correct
 
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