Question

Help with question 6 pleasee! :) http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w05_qp_3.pdf

Answer 1

@ganeshie8 @hartnn

Answer 2

HI!!
i guess we can do 6, hold on one second

Answer 3

HII. Okay! :))

Answer 4

let me write it with pencil first, not sure if it will work on the fly

Answer 5

Okayy. :) Take your time.

Answer 6

oh yeah you do just what is says, get it in one step more or less

Answer 7

\[x=\sin^2(\theta), dx =2\sin(\theta)\cos(\theta)d\theta\] gives
\[2\int\sqrt{\frac{\sin^2(\theta)}{1-\sin^2(\theta)}}\sin(\theta)\cos(\theta)d\theta\]

Answer 8

using \(1-\sin^2(\theta)=\cos^2(\theta)\) and cancelling you get what they wrote in one step

Answer 9

oh oh, maybe i said too much @iGreen is this ok, or should i ask "what do you think?" at this step?

Answer 10

im quite confused actually. where do i begin from?

Answer 11

ok lets back up
it says make \(x=\sin^2(\theta)\) right?

Answer 12

so that is what i did
i says "let \(x=\sin^2(\theta)\)" then we need \(dx\)

Answer 13

yes yes

Answer 14

so we get
\[dx=2\sin(\theta)\cos(\theta)\] right?

Answer 15

why 2sinxcosx?

Answer 16

on account of that is the derivative of \(\sin^2(\theta)\)

Answer 17

via the chain rule

Answer 18

clear or no?

Answer 19

ohh okok

Answer 20

so now substitute exactly as you see it \[2\int\sqrt{\frac{\sin^2(\theta)}{1-\sin^2(\theta)}}\sin(\theta)\cos(\theta)d\theta\]

Answer 21

then use \[1-\sin^2(\theta)=\cos^2(\theta)\]

Answer 22

\[2\int\sqrt{\frac{\sin^2(\theta)}{\cos^2(\theta)}}\sin(\theta)\cos(\theta)d\theta\]

Answer 23

then cancel your brains out and end up with
\[2\int \sin^2(\theta)d\theta\]

Answer 24

nicee. thankss.

Answer 25

nicee. thankss.

Answer 26

\[\color\magenta\heartsuit\]