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you got it or you want me to finish the whole thing
nope ....... its fine ....... and thank you very much .....
have some more
and thanks again
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you got it or you want me to finish the whole thing
we have tto find value of w?y
yeap its pure
i just hate ....hate this chapter -_-
i never get the answer
oops sorrywe have tto find value of w?
and then how wud i sketch :0 lemme try
you're welcomenope ....... its fine ....... and thank you very much .....
have some more
and thanks again
i dont remember sorryoops sorry
just do b part 2
its okayi dont remember sorry
ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=ohttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
applepie1996 or Rutzaba
its question 7 (i) ....i so hate this topic and the complex one
Silent Hunterok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2
x= pi/4
so the limits be pi/4 and 0
by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx
now substituting
(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24
No need to thank me
because i didn't solve it
Rutzaba solved it for me
lucky usRutzaba and applepie1996
thank you very much ..... ( there maybe more questions on the way )
and thanks again
awwwwwlyt jane wali in 2 mins... rat tak post krdungi ans insha Allah
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