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20 21 Differentiation and Integration of Trigonometry Function 9709

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π‘₯2
𝑑𝑦
1. (a) Show that 𝑑π‘₯ (π‘₯ − tan−1 π‘₯) = 1+π‘₯ 2 .
2
√3
[2]
1
(i) Show that ∫0 π‘₯ tan−1 π‘₯ 𝑑π‘₯ = 3 πœ‹ − 2 √3.
[5]
9709/03/SP/20/# 5
1
1
2. The variables π‘₯ and 𝑦 satisfy the relation sin 𝑦 = tan π‘₯, where − 2 πœ‹ < 𝑦 < 2 πœ‹. Show
that
𝑑𝑦
1
=
𝑑π‘₯ cos π‘₯ √cos 2π‘₯
[5]
9709/32/F/M/19/# 5
1
3. (i) Differentiate sin2 πœƒ with respect to πœƒ.
[2]
−𝟐 𝐜𝐨𝐬 𝜽
−𝟐 𝐬𝐒𝐧 𝜽 𝐜𝐨𝐬 𝜽
−𝟐 𝐜𝐨𝐬𝐞𝐜 𝟐 𝜽 𝐜𝐨𝐭 𝜽 , 𝒐𝒓
𝒐𝒓
πŸ‘
𝐬𝐒𝐧 𝜽
π¬π’π§πŸ’ 𝜽
(ii) The variables π‘₯ and πœƒ satisfy the differential equation
𝑑π‘₯
π‘₯ tan πœƒ
+ cosec 2 πœƒ = 0,
π‘‘πœƒ
1
1
for 0 < πœƒ < 2 πœ‹ and π‘₯ > 0. It is given that π‘₯ = 4 when πœƒ = 6 πœ‹. Solve the
differential equation, obtaining an expression for π‘₯ in terms of πœƒ.
[6]
𝒙 = √𝐜𝐨𝐬𝐞𝐜 𝟐 𝜽 + 𝟏𝟐
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1
4. The diagram shows the curve 𝑦 = sin 3π‘₯ cos π‘₯ for 0 ≤ π‘₯ ≤ 2 πœ‹ and its minimum point
𝑀. The shaded region 𝑅 is bounded by the curve and the π‘₯-axis.
(i) By expanding sin(3π‘₯ + π‘₯) and sin(3π‘₯ − π‘₯) show that
1
sin 3π‘₯ cos π‘₯ = 2 (sin 4π‘₯ + sin 2π‘₯)
[3]
(ii) Using the result of part (i) and showing all necessary working, find the exact area of
region 𝑅.
[4]
πŸ—
πŸπŸ”
𝑑𝑦
(iii)Using the result of part (i), express 𝑑π‘₯ in terms of cos 2π‘₯ and hence find the π‘₯coordinate of 𝑀, giving your answer correct to 2 decimal places.
[5]
𝟏. πŸπŸ—
9709/32/M/J/19/#10
1
1
5. The curve 𝑦 = sin (π‘₯ + 3 πœ‹) cos π‘₯ has two stationary points in the interval 0 ≤ π‘₯ ≤ πœ‹.
(i)
(ii)
(iii)
6. (i)
(ii)
𝑑𝑦
Find 𝑑π‘₯ .
[2]
𝟏
𝐜𝐨𝐬 (πŸπ’™ + 𝝅)
πŸ‘
By considering the formula for cos(𝐴 + 𝐡), show that, at the stationary points on
1
the curve, cos (2π‘₯ + 3 πœ‹) = 0.
[2]
Hence find the exact π‘₯-coordinate of the stationary points.
[3]
𝟏
πŸ•
𝝅 𝒐𝒓
𝝅
𝟏𝟐
𝟏𝟐
9709/33/M/J/19/#7
By differentiating
1
πœ‹
2
1
πœ‹
4
cos π‘₯
sin π‘₯
𝑑𝑦
, show that if 𝑦 = cot π‘₯ then 𝑑π‘₯ = − 𝐜𝐨𝐬𝐞𝐜 𝟐 𝒙.
1
Show that ∫ π‘₯ cosec 2 π‘₯ 𝑑π‘₯ = 4 (πœ‹ + ln 4).
[2]
[6]
9709/31/O/N/19/#6
7. The variable π‘₯ and πœƒ satisfy the differential equation
1 𝑑π‘₯
1
sin πœƒ
= (π‘₯ + 2) cos πœƒ
2 π‘‘πœƒ
2
1
for 0 < πœƒ < πœ‹. It is given that π‘₯ = 1 when πœƒ = 3 πœ‹. Solve the differential equation and
obtain an expression for π‘₯ in terms of cos πœƒ.
[8]
𝒙 = πŸ’ − πŸ” 𝒄𝒐𝒔 𝜽
9709/32/O/N/19/#6
8. The diagram shows the graph of 𝑦 = 𝑒 cos π‘₯ sin3 π‘₯ for 0 ≤ π‘₯ ≤ πœ‹, and its maximum point
𝑀. The shaded region 𝑅 is bounded by the curve and the π‘₯-axis.
(i)
(ii)
9. (i)
Find the π‘₯-coordinate of 𝑀. Show all necessary working and give your answer
correct to 2 decimal places.
[5]
𝒙 = 𝟏. πŸπŸ”
By first using the substitution 𝑒 = cos π‘₯, find the exact of the area of 𝑅.
[7]
πŸ’
𝒆
9709/33/O/N/19/#10
Using the expansions of cos(3π‘₯ + π‘₯) and cos(3π‘₯ − π‘₯), show that
1
(cos 4π‘₯ + cos 2π‘₯) ≡ cos 3π‘₯ cos π‘₯
2
[3]
(ii)
1
πœ‹
6
1
− πœ‹
6
Hence show that ∫
3
cos 3π‘₯ cos π‘₯ 𝑑π‘₯ = 8 √3.
[3]
9709/32/F/M/18/#3
2
10. The variable π‘₯ and πœƒ satisfy the differential equation
𝑑π‘₯
π‘₯ sin2 πœƒ
= 2 tan πœƒ + 1,
π‘‘πœƒ
1
1
for 0 ≤ πœƒ ≤ 2 πœ‹ and π‘₯ > 0. It is given that π‘₯ = 1 when πœƒ = 4 πœ‹.
(i)
(ii)
𝑑
Show that π‘‘πœƒ (tan2 πœƒ) =
2 tan πœƒ
cos2 πœƒ
.
[1]
1
Solve the differential equation and calculate the value of π‘₯ when πœƒ = 3 πœ‹, giving
your answer correct to 3 significant figures.
[7]
𝒙 = 𝟐. πŸ“πŸ’
9709/32/F/M/18/#6
11. A curve has equation 𝑦 =
𝑒 3π‘₯
1
2
tan π‘₯
. Find the π‘₯-coordinates of the stationary points of the
curve in the interval 0 < πœƒ < πœ‹. Give your answer correct to 3 decimal places
[6]
𝟎. πŸ‘πŸ’πŸŽ 𝒐𝒓 𝟐. πŸ–πŸŽπŸ
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3
4
1
4
π‘₯
12. Let that 𝐼 = ∫ √1−π‘₯ 𝑑π‘₯.
1
πœ‹
3
1
πœ‹
6
2
(i)
Using the substitution that π‘₯ = cos πœƒ, show that that 𝐼 = ∫ 2 cos 2 πœƒ π‘‘πœƒ.
(ii)
Hence find the exact value of 𝐼.
(iv)
Find 𝑑π‘₯ .
[4]
[4]
𝟏
𝝅
πŸ”
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1
13. The curve 𝑦 = sin (π‘₯ + 3 πœ‹) cos π‘₯ has two stationary points in the interval 0 ≤ π‘₯ ≤ πœ‹.
(v)
(vi)
14. (i)
(iii)
𝑑𝑦
[2]
𝟏
𝐜𝐨𝐬 (πŸπ’™ + 𝝅)
πŸ‘
By considering the formula for cos(𝐴 + 𝐡), show that, at the stationary points on
1
the curve, cos (2π‘₯ + 3 πœ‹) = 0.
[2]
Hence find the exact π‘₯-coordinate of the stationary points.
[3]
𝟏
πŸ•
𝝅 𝒐𝒓
𝝅
𝟏𝟐
𝟏𝟐
9709/33/M/J/19/#7
By differentiating
1
πœ‹
2
1
πœ‹
4
cos π‘₯
𝑑𝑦
, show that if 𝑦 = cot π‘₯ then 𝑑π‘₯ = − 𝐜𝐨𝐬𝐞𝐜 𝟐 𝒙.
sin π‘₯
1
Show that ∫ π‘₯ cosec 2 π‘₯ 𝑑π‘₯ = 4 (πœ‹ + ln 4).
[2]
[6]
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15. The variable π‘₯ and πœƒ satisfy the differential equation
1 𝑑π‘₯
1
sin πœƒ
= (π‘₯ + 2) cos πœƒ
2 π‘‘πœƒ
2
1
for 0 < πœƒ < πœ‹. It is given that π‘₯ = 1 when πœƒ = 3 πœ‹. Solve the differential equation and
obtain an expression for π‘₯ in terms of cos πœƒ.
[8]
𝒙 = πŸ’ − πŸ” 𝒄𝒐𝒔 𝜽
9709/32/O/N/19/#6
3
16. The diagram shows the graph of 𝑦 = 𝑒 cos π‘₯ sin3 π‘₯ for 0 ≤ π‘₯ ≤ πœ‹, and its maximum point
𝑀. The shaded region 𝑅 is bounded by the curve and the π‘₯-axis.
(iii) Find the π‘₯-coordinate of 𝑀. Show all necessary working and give your answer
correct to 2 decimal places.
[5]
𝒙 = 𝟏. πŸπŸ”
(iv)
By first using the substitution 𝑒 = cos π‘₯, find the exact of the area of 𝑅.
[7]
πŸ’
𝒆
9709/33/O/N/19/#10
4
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