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Integration using trigonometric identities questions with full Worked solutions
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  • Course
  • Calculus
  • Institution
  • University Of East London (UEL)

This document will allow university Math students to practice integration using trigonometric identities. All the questions are similar in style and full worked solutions have been provided. This will allow you to understand how a certain question in this document is answered and then you will be a...

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Document information

  • November 13, 2024
  • 27
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

Subjects

  • calculus
  • trigonometric identities
  • integration

Written for

  • University of East London (UEL)
  • Unknown
  • Calculus
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Integration using trigonometric identities

2 4 7 4
1) Calculate ∫ 5
𝑠𝑖𝑛 𝑥 − 8
𝑐𝑜𝑠 𝑥 𝑑𝑥

2 1 1
𝑠𝑖𝑛 𝑥 = 2
− 2
𝑐𝑜𝑠2𝑥
4 1 1 2
𝑠𝑖𝑛 𝑥 = ( 2 − 2
𝑐𝑜𝑠2𝑥)

2 1 1
𝑐𝑜𝑠 𝑥 = 2
+ 2
𝑐𝑜𝑠2𝑥
4 1 1 2
𝑐𝑜𝑠 𝑥 = ( 2 + 2
𝑐𝑜𝑠2𝑥)

2 1 1 2 7 1 1 2
∫ 5
(2 − 2
𝑐𝑜𝑠2𝑥) − 8
(2 + 2
𝑐𝑜𝑠2𝑥) 𝑑𝑥
2 1 1 1 1 2 7 1 1 1 1 2
= ∫ 5
(4 − 4
𝑐𝑜𝑠2𝑥 − 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) − 8
(4 + 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) 𝑑𝑥
2 1 1 1 2 7 1 1 1 2
= ∫ 5
(4 − 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) − 8
(4 + 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) 𝑑𝑥
1 1 1 2 7 7 7 2
= ∫ 10
− 5
𝑐𝑜𝑠2𝑥 + 10
𝑐𝑜𝑠 2𝑥 − 32
− 16
𝑐𝑜𝑠2𝑥 − 32
𝑐𝑜𝑠 2𝑥 𝑑𝑥
19 51 19 2
= ∫− 160
− 80
𝑐𝑜𝑠2𝑥 − 160
𝑐𝑜𝑠 2𝑥 𝑑𝑥

2 1 1
𝑐𝑜𝑠 𝑥 = 2
+ 2
𝑐𝑜𝑠2𝑥
2 1 1
𝑐𝑜𝑠 2𝑥 = 2
+ 2
𝑐𝑜𝑠4𝑥

19 51 19 1 1
= ∫− 160
−
80
𝑐𝑜𝑠2𝑥 − 160 (2 + 2
𝑐𝑜𝑠4𝑥) 𝑑𝑥
19 51 19 19
= ∫− 160
− 80 𝑐𝑜𝑠2𝑥 − 320 − 320
𝑐𝑜𝑠4𝑥 𝑑𝑥
57 51 19
= ∫− 320
− 80 𝑐𝑜𝑠2𝑥 − 320 𝑐𝑜𝑠4𝑥 𝑑𝑥
57 51 19
=− 320
𝑥 − 160 𝑠𝑖𝑛2𝑥 − 1280 𝑠𝑖𝑛4𝑥 + 𝐶


FINAL ANSWER

57 51 19
=− 320
𝑥 − 160
𝑠𝑖𝑛2𝑥 − 1280
𝑠𝑖𝑛4𝑥 + 𝐶

, 3 4 2 4
2) Calculate ∫ 7
𝑠𝑖𝑛 𝑥 − 9
𝑐𝑜𝑠 𝑥 𝑑𝑥

2 1 1
𝑠𝑖𝑛 𝑥 = 2
− 2
𝑐𝑜𝑠2𝑥
4 1 1 2
𝑠𝑖𝑛 𝑥 = ( 2 − 2
𝑐𝑜𝑠2𝑥)

2 1 1
𝑐𝑜𝑠 𝑥 = 2
+ 2
𝑐𝑜𝑠2𝑥
4 1 1 2
𝑐𝑜𝑠 𝑥 = ( 2 + 2
𝑐𝑜𝑠2𝑥)

3 1 1 2 2 1 1 2
= ∫ 7
(2 − 2
𝑐𝑜𝑠2𝑥) − 9
(2 + 2
𝑐𝑜𝑠2𝑥) 𝑑𝑥
3 1 1 1 1 2 2 1 1 1 1 2
= ∫ 7
(4 − 4
𝑐𝑜𝑠2𝑥 − 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) − 9
(4 + 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) 𝑑𝑥
3 1 1 1 2 2 1 1 1 2
= ∫ 7
(4 − 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) − 9
(4 + 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) 𝑑𝑥
3 3 3 2 1 1 1 2
= ∫ 28
− 14
𝑐𝑜𝑠2𝑥 + 28
𝑐𝑜𝑠 2𝑥 − 18
− 9
𝑐𝑜𝑠2𝑥 − 18
𝑐𝑜𝑠 2𝑥 𝑑𝑥
13 41 13 2
= ∫ 252
− 126
𝑐𝑜𝑠2𝑥 + 252
𝑐𝑜𝑠 2𝑥 𝑑𝑥

2 1 1
𝑐𝑜𝑠 𝑥 = 2
+ 2
𝑐𝑜𝑠2𝑥
2 1 1
𝑐𝑜𝑠 2𝑥 = 2
+ 2
𝑐𝑜𝑠4𝑥

13 41 13 1 1
= ∫ 252
− 126
𝑐𝑜𝑠2𝑥 + 252
( 2 + 2 𝑐𝑜𝑠4𝑥) 𝑑𝑥
13 41 13 13
= ∫ 252 − 126
𝑐𝑜𝑠2𝑥 + 504
+ 504 𝑐𝑜𝑠4𝑥 𝑑𝑥
13 41 13
= ∫ 168 − 126
𝑐𝑜𝑠2𝑥 + 504
𝑐𝑜𝑠4𝑥 𝑑𝑥
13 41 13
= 168
𝑥 − 252
𝑠𝑖𝑛2𝑥 + 2016
𝑠𝑖𝑛4𝑥 + 𝐶


FINAL ANSWER

13 41 13
= 168
𝑥− 252
𝑠𝑖𝑛2𝑥 + 2016
𝑠𝑖𝑛4𝑥 + 𝐶

, 4 4
3) Calculate ∫5𝑠𝑖𝑛 𝑥 − 8𝑐𝑜𝑠 𝑥 𝑑𝑥

2 1 1
𝑠𝑖𝑛 𝑥 = 2
− 2
𝑐𝑜𝑠2𝑥
4 1 1 2
𝑠𝑖𝑛 𝑥 = ( 2 − 2
𝑐𝑜𝑠2𝑥)

2 1 1
𝑐𝑜𝑠 𝑥 = 2
+ 2
𝑐𝑜𝑠2𝑥
4 1 1 2
𝑐𝑜𝑠 𝑥 = ( 2 + 2
𝑐𝑜𝑠2𝑥)

1 1 2 1 1 2
∫5( 2 − 2
𝑐𝑜𝑠2𝑥) − 8( 2 + 2
𝑐𝑜𝑠2𝑥) 𝑑𝑥
1 1 2 1 1 2
= ∫5( 2 − 2
𝑐𝑜𝑠2𝑥) − 8( 2 + 2
𝑐𝑜𝑠2𝑥) 𝑑𝑥
1 1 1 1 2 1 1 1 1 2
= ∫5( 4 − 4
𝑐𝑜𝑠2𝑥 − 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) − 8( 4 + 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) 𝑑𝑥
1 1 1 2 1 1 1 2
= ∫5( 4 − 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) − 8( 4 + 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥) 𝑑𝑥
5 5 5 2 2
= ∫ 4
− 2
𝑐𝑜𝑠2𝑥 + 4
𝑐𝑜𝑠 2𝑥 − 2 − 4𝑐𝑜𝑠2𝑥 − 2𝑐𝑜𝑠 2𝑥 𝑑𝑥
3 13 3 2
= ∫− 4
− 2
𝑐𝑜𝑠2𝑥 − 4
𝑐𝑜𝑠 2𝑥 𝑑𝑥

2 1 1
𝑐𝑜𝑠 𝑥 = 2
+ 2
𝑐𝑜𝑠2𝑥
2 1 1
𝑐𝑜𝑠 2𝑥 = 2
+ 2
𝑐𝑜𝑠4𝑥

3 13 3 1 1
= ∫− 4 2
−
𝑐𝑜𝑠2𝑥 − 4 ( 2 + 2 𝑐𝑜𝑠4𝑥) 𝑑𝑥
3 13 3 3
= ∫− 4
− 2 𝑐𝑜𝑠2𝑥 − 8 − 8 𝑐𝑜𝑠4𝑥 𝑑𝑥
9 13 3
= ∫− 8
− 2 𝑐𝑜𝑠2𝑥 − 8 𝑐𝑜𝑠4𝑥 𝑑𝑥
9 13 3
=− 8
𝑥 − 4 𝑠𝑖𝑛2𝑥 − 32 𝑠𝑖𝑛4𝑥 + 𝐶


FINAL ANSWER

9 13 3
=− 8
𝑥− 4
𝑠𝑖𝑛2𝑥 − 32
𝑠𝑖𝑛4𝑥 + 𝐶

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