Edexcel A2 C4 Mathematics June 2016 - Official Thread

Would you get the marks for this I never see it mentioned on the examiner reports or markschemes?

How do you integrate this?
Thanks for the help!


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Hey guys, this is a really dum dum question but what does it mean, with regards to the dot product rule, when two directional vectors are parallel. Also does a and b in the dot product rule HAVE to be directional vectors because I think I saw a question where they used a position vector 0.O... i was probably imagining it.

It means a.b is just the product of the moduli/lengths of the vectors.

Thanks, and do a and b have to be directional vectors or can they be position vectors?

Substitution

What shall I substitute?


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x = sec u - 1

x = sec u - 1

x = sec u - 1

How did you figure out that was what you were meant to substitute? Why that particular trig function?


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I get down to $\int \sec u \tan^2 u \mathrm{d}u$

Not sure how to integrate that?

Btw how could I get a space between the second u and the du?

$x^2 + 2x \equiv (x+1)^2 - 1$ make the sub $u = x+1$ to get $\int \sqrt{u^2 - 1}$ which is just begging for a $u = \sec v$ sub, so I combined them into one and gave it to you.

Be warned that this is an incredibly tedious and boring integral whose antiderivative is supremely ugly and will likely take you an entire line to write.

$\sec^3 u + \sec u$ then the latter is standard and IBP and/or reduction formulae is needed for the first part.



whatever \, \mathrm{d}u

$x^2 + 2x \equiv (x+1)^2 - 1$ make the sub $u = x+1$ to get $\int \sqrt{u^2 - 1}$ which is just begging for a $u = \sec v$ sub, so I combined them into one and gave it to you.

Be warned that this is an incredibly tedious and boring integral whose antiderivative is supremely ugly and will likely take you an entire line to write.

They won't give an integral like this in a C4 paper, right?


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