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Title: Cambridge International Examination,General certificate of Education,Advanced Subsidiary level P2 Pure maths, October/November,university Of Cambridge Local Examination Syndicate
Description: Cambridge International Examination General Certificate of Education Advanced Subsidiary Level Mathematics Paper 2 Pure mathematics (P2) OCTOBER / NOVEMBER SESSION 2001 University Of Cambridge Local Examination Syndicate
Description: Cambridge International Examination General Certificate of Education Advanced Subsidiary Level Mathematics Paper 2 Pure mathematics (P2) OCTOBER / NOVEMBER SESSION 2001 University Of Cambridge Local Examination Syndicate
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Cambridge International Examination
General Certificate of Education Advanced Subsidiary Level
Mathematics
Paper 2 Pure mathematics (P2)
OCTOBER / NOVEMBER SESSION 2001
University Of Cambridge
Local Examination Syndicate
Show that β«
Q7(i)
π
π
4
1
4
and that β«
2
0
sin2x dx =
0
(ii) Use the result in part (i) to evaluate
Now β«
= 2β«
π
4
0
0
π
4
(2sinx + 3cosx) 2 dx
0
Solution (i)
Since sin2x = 2sinxcosx
d
Also
sinx = cosx
dx
d(sinx) = cosx dx
sin2xdx = β«
π
4
β«
sinx d(sinx) = 2
π
4
0
2sinxcosx dx = 2β«
π
π
4
0
sinx (cosxdx)
π
4
(sinx) 2
2
1
cos 2 x dx = (π + 2)
8
= (sinx) 2
4
0
= sin
0
π
2
4
-
(sin0) 2
2
2
=
-0 =
2
1
2
π
1
2
Also cos2x = cos x - sin 2 x ------------ (1)
sin 2 x + cos 2 x = 1
βΉ sin 2 x = 1 - cos 2 x ------------ (2)
From equation (1) and equation (2)
cos2x = cos 2 x - 1 - cos 2 x = cos 2 x - 1 + cos 2 x = 2cos 2 x - 1
Hence β«
4
0
2
sin2x dx =
cos2x + 1 = 2 cos 2 x
1 + cos2x
= cos 2 x
2
π
π
π
π
1 4
1 4
4
4 1 + cos2x
βΉβ«
cos 2 xdx = β«
dx = β« dx + β« cos2x dx
0
0
2
2 0
2 0
π
π
1
1 sin2x
= [x] 04 +
2
2
2
=
π
8
+
4
=
0
1 π 1
π
+
sin2
- sin2(0)
2 4 4
4
1
π
π 1
π 2
sin - sin(0) =
+ ( 1) =
+
4
2
8 4
8 8
βΉβ«
π
4
0
1
cos 2 x dx = ( π + 2)
8
(ii)
I=β«
π
4
0
(2sinx + 3cosx) 2 dx
------------ (1)
Now (2sinx + 3cosx) 2 = 4sin 2 x + 9cos 2 x + 12sinxcosx
= 4sin 2 x + 9cos 2 x + 6 (2sinxcosx) --------- (2)
Also
sin 2 x + cos 2 x = 1
2
βΉ sin x = 1 - cos 2 x ----------- (3)
2sinxcosx = sin2x --------------- (4)
Putting equation (3) & equation (4) in equation (2)
(2sinx + 3cosx) 2 = 4 1 - cos 2 x + 9 cos 2 x + 6 sin2x
= 4 - 4cos 2 x + 9cos 2 x + 6sin2x
= 4 + 5cos 2 x + 6sin2x ---------- (5)
Putting equation (5) in equation (1)
I=β«
π
4
0
4 + 5cos 2 x + 6sin2x dx
= 4β«
π
4
0
dx + 5β«
π
=4
= π + 5β«
Putting β«
I=β«
4
π
4
0
+ 5β«
0
0
4
0
π
4
0
cos 2 x dx + 6β«
cos 2 x dx + 6 β«
cos 2 x dx + 6β«
cos 2 x dx + 6β«
π
4
4
π
π
= 4 [x] 04 + 5β«
π
4
4
0
4
0
sin2x dx
π
4
0
4
0
sin2x dx
π
sin2x dx
sin2x dx ----------- (6)
π
cos 2 x dx =
1
1
4
(π + 2) andβ« sin2xdx = in equation (6)
8
0
2
π
0
π
π
(2sinx + 3cosx) 2 dx = π + 5
=
1
1
5π 10
( π + 2) + 6
= π+
+
+3
8
2
8
8
13π 34
+
8
8
Title: Cambridge International Examination,General certificate of Education,Advanced Subsidiary level P2 Pure maths, October/November,university Of Cambridge Local Examination Syndicate
Description: Cambridge International Examination General Certificate of Education Advanced Subsidiary Level Mathematics Paper 2 Pure mathematics (P2) OCTOBER / NOVEMBER SESSION 2001 University Of Cambridge Local Examination Syndicate
Description: Cambridge International Examination General Certificate of Education Advanced Subsidiary Level Mathematics Paper 2 Pure mathematics (P2) OCTOBER / NOVEMBER SESSION 2001 University Of Cambridge Local Examination Syndicate