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RD Sharma - Volume 1

19. Indefinite Integrals

Exercise 19.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

Exercise 19.3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Exercise 19.4

Exercise 19.5

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Exercise 19.6

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Exercise 19.7

Exercise 19.8

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Exercise 19.9

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72

Exercise 19.10

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Exercise 19.11

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Exercise 19.12

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Exercise 19.13

Exercise 19.14

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Exercise 19.15

Exercise 19.16

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Exercise 19.17

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Exercise 19.18

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Exercise 19.19

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Exercise 19.20

1 2 3 4 5 6 7 8 9 10

Exercise 19.21

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Exercise 19.22

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Exercise 19.23

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Exercise 19.24

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Exercise 19.25

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Exercise 19.26

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Exercise 19.27

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Exercise 19.28

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Exercise 19.29

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Exercise 19.30

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Exercise 19.31

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Exercise 19.32

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Very short answer

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62

MCQ

18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35

Revision exercise

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Q. 10

Evaluate <span la

Class 12th RD Sharma - Volume 1 19. Indefinite Integrals

Answer :

∫cosec²x(cos²x-sin²x)²dx

Opening the square

On multiplying (1-sin²x)(1-sin²x) equation reduces to

=∫(cosec²x-2+sin²x-2cos²x+ sin²x)dx

=∫(cosec²x-2+2sin²x-2cos²x)dx

=∫(-2(cos²x-sin²x)+cosec²x-2)dx

=∫(-2cos2x+cosec²x-2)dx

On solving this we get our answer i.e

=-sin2x-cotx-2x+c

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