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Edexcel C3,C4 June 2013 Thread

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lol that question is on Q3 page 140 not 142

i got the older version of the book, so there is only 1 page on exam style questions and i guess it was a coincidence that question 3 was substitution too, so i automatically thought she was asking about that question.

Reply 1121

masryboy94

Original post by nukethemaly

Oh yeah and I've already done that one. It's just the trig substitutions I have trouble with.

what was the answer?

Reply 1122

usycool1

Original post by nukethemaly

Yeah that one! I've attached my neatest try! and also the question

EDIT: I literally don't know what i'm trying to do with those limits there haha

That's a nice question, plenty of trigonometry in it. :tongue:

OK, firstly, you got

\dfrac{dx}{d\theta}

. There's no need to write it as

\dfrac{d\theta}{dx}

in this case; just substitute in:

\sec \theta \tan \theta d\theta

\sec^2 \theta

and don't forget to change the limits.

So:

\displaystyle \int_2^3 \dfrac{1}{(x^2-1)^{\frac{3}{2}}} dx

Write it as:

\displaystyle \int_\frac{\pi}{3}^{arccos(\frac{1}{3})} \dfrac{\sec \theta \tan \theta}{(\sec^2 \theta -1)^{\frac{3}{2}}} d \theta

Now, how else can you write

sec^2 \theta - 1

? Then do some simplifying and you should get it. :smile:

(edited 10 years ago)

Reply 1123

usycool1

Original post by masryboy94

what was the answer?

I'm getting 0.0940 (to 4 s.f). :smile:

Reply 1124

masryboy94

Original post by usycool1

That's a nice question, plenty of trigonometry in it. :tongue:

OK, firstly, you got

\dfrac{dx}{d\theta}

. There's no need to write it as

\dfrac{d\theta}{dx}

in this case; just substitute in

\sec \theta \tan \theta d\theta

\sec^2 \theta

and don't forget to change the limits.

So:

\displaystyle \int_2^3 \dfrac{1}{(x^2-1)^{\frac{3}{2}}} dx

Write it as:

\displaystyle \int_\frac{\pi}{3}^{arccos(\frac{1}{3})} \dfrac{\sec \theta \tan \theta}{(\sec^2 \theta -1)^{\frac{3}{2}}} d \theta

Now, how else can you write

sec^2 \theta - 1

? Then do some simplifying and you should get it. :smile:

Original post by usycool1

I'm getting 0.0940 (to 4 s.f). :smile:

haha you beat me to it, yep got that answer ! just like you said :biggrin:

hope you get it now Aly :biggrin:

Reply 1125

masryboy94

Original post by usycool1

I'm getting 0.0940 (to 4 s.f). :smile:

out of curiosity how did you simplify from secxtanx/tan^3 x ? because it took me a good while to try simplify it and there must of been an easier way.

Reply 1126

nukethemaly

Original post by usycool1

That's a nice question, plenty of trigonometry in it. :tongue:

OK, firstly, you got

\dfrac{dx}{d\theta}

. There's no need to write it as

\dfrac{d\theta}{dx}

in this case; just substitute in:

\sec \theta \tan \theta d\theta

\sec^2 \theta

and don't forget to change the limits.

So:

\displaystyle \int_2^3 \dfrac{1}{(x^2-1)^{\frac{3}{2}}} dx

Write it as:

\displaystyle \int_\frac{\pi}{3}^{arccos(\frac{1}{3})} \dfrac{\sec \theta \tan \theta}{(\sec^2 \theta -1)^{\frac{3}{2}}} d \theta

Now, how else can you write

sec^2 \theta - 1

? Then do some simplifying and you should get it. :smile:

holy ****, it's that easy!

Reply 1127

nukethemaly

Original post by masryboy94

haha you beat me to it, yep got that answer ! just like you said :biggrin:

hope you get it now Aly :biggrin:

Yep I do!

Thanks you guys!

Reply 1128

Kvothe the Arcane

Original post by masryboy94

out of curiosity how did you simplify from secxtanx/tan^3 x ? because it took me a good while to try simplify it and there must of been an easier way.

I haven't looked at what he's done, but I'd simplify like this:

\dfrac{sec(x)tan(x)}{tan^3(x)}

\dfrac{sec(x)tan(x)}{tan(x)(tan^2(x))}

\dfrac{sec(x)}{sec^2(x)-1}

(edited 10 years ago)

Reply 1129

masryboy94

Original post by reubenkinara

I haven't looked at what he's done, but I'd simplify like this:

\dfrac{sec(x)tan(x)}{tan^3(x)}

\dfrac{sec(x)tan(x)}{tan(x)(tan^2(x))}

\dfrac{sec(x)}{sec^2(x)-1}

but then how would you integrate that?

Reply 1130

Kvothe the Arcane

Original post by masryboy94

but then how would you integrate that?

Jut realized the context. :eek:

Reply 1131

masryboy94

i've just been doing a soloman paper now and on the mark scheme to a bionomial question they gave the answer as follows: (attachment). my problem is i got the coefficient of x³to be 160 however they got -160. but if you were to look at their calculation, its should be (4x8) - (2 x -64) which is 160. so have they done a mistake there with the sign? or am i not catching on something?

(edited 10 years ago)

Screen Shot 2013-05-09 at 00.11.46.png9.5KB

Reply 1132

Proflash

Original post by masryboy94

but then how would you integrate that?

i tried to sub in x=sec and attempted the f'(x)/f(x) integration but im getting it wrong for some reason.

Reply 1133

thelion0

Original post by masryboy94

I did that paper today. The mark-scheme is wrong.

Reply 1134

MathMan

Rather un-intuitive but use the substitution u=tanx

Reply 1135

Lanre

Original post by masryboy94

but then how would you integrate that?

Wouldn't it be -(sec^2 (x)-1)^(-0.5) if you diffentiate that you should get that function.

(edited 10 years ago)

Reply 1136

Vaner

Hey!

I'm stuck on question 3 and 8 in the C4 specimen paper. I'm in the early stages of solving and would really appreciate if any solves them for me as they are not in ExamSolutions.

Thank you in advance!

Reply 1137

justinawe

Original post by Vaner

I'm going out to the movies right now (Iron Man 3 :cool:

), but if no one's helped you out when I get back, I'll take a look at it.

Reply 1138

usycool1

Original post by Vaner

The integration by substitution and differential equation question? What have you tried? :smile:

Original post by masryboy94

out of curiosity how did you simplify from secxtanx/tan^3 x ? because it took me a good while to try simplify it and there must of been an easier way.

I simplified it down to -cosec x :tongue:

Original post by Lanre

Wouldn't it be -(sec^2 (x)-1)^(-0.5) if you diffentiate that you should get that function.

Nope, try differentiating it. :smile:

Reply 1139

masryboy94

Original post by usycool1

I simplified it down to -cosec x :tongue:

ooo so you got to the point cosx/sin^2 x and then brought the sin^2 x to the top, did the reverse chain rule to get -1/sinx hence getting the -cosecx, right? :smile: