# Misc 27 - Chapter 7 Class 12 Integrals

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 27 Evaluate the definite integral 0𝜋2cos2𝑥 𝑑𝑥cos2𝑥 + 4sin2𝑥 Let I = 0𝜋2𝑐𝑜𝑠2𝑥𝑐𝑜𝑠2𝑥 + 4𝑠𝑖𝑛2𝑥 𝑑𝑥 = 0𝜋2𝑐𝑜𝑠2𝑥𝑐𝑜𝑠2𝑥 + 4 1 − 𝑐𝑜𝑠2𝑥 𝑑𝑥 = 0𝜋2𝑐𝑜𝑠2𝑥4 − 3 𝑐𝑜𝑠2𝑥 𝑑𝑥 = −13 0𝜋4 −3 𝑐𝑜𝑠2 𝑥 4 − 3 𝑐𝑜𝑠2𝑥 𝑑𝑥 = −130𝜋24 −3 𝑐𝑜𝑠2𝑥 −44 − 3 𝑐𝑜𝑠2𝑥 𝑑𝑥 = −130𝜋21−44 − 3 𝑐𝑜𝑠2𝑥 𝑑𝑥 = −130𝜋2𝑑𝑥+430𝜋2𝑑𝑥4 − 3 𝑐𝑜𝑠2𝑥 Dividing numerator and denominator by 𝑐𝑜𝑠2 𝑥 = −13𝜋2+43 0𝜋2𝑑𝑥𝑐𝑜𝑠2 𝑥4 − 3 𝑐𝑜𝑠2𝑥 𝑐𝑜𝑠2 𝑥 𝑑𝑥 = −13𝜋2+43 0𝜋2𝑠𝑒𝑐2𝑥4 𝑠𝑒𝑐2𝑥 − 3 𝑑𝑥 = −𝜋6+43 0𝜋2𝑠𝑒𝑐2𝑥4 (1 + 𝑡𝑎𝑛2𝑥) − 3 𝑑𝑥 Put tan x = t so that 𝑠𝑒𝑐2 x dx = dt Thus, When x = 0, t = 0, and when x = 𝜋2, 𝑡= Substituting values and limit ∴ I =−𝜋6+43 0uc1𝑑𝑡4 1+ 𝑡2 −3 =−𝜋6+43 0uc1𝑑𝑡4𝑡2+1 =−𝜋6+43 ∙14 0uc1𝑑𝑡 𝑡2+14 =𝜋6+43∙14 × 112 𝑡𝑎𝑛−1𝑡120uc1 = −𝜋6+23∙ 𝑡𝑎𝑛−1 2𝑡0uc1 =−𝜋6+23∙[𝑡𝑎𝑛−1uc1−𝑡𝑎𝑛−10] = −𝜋6+23∙𝜋2−0 =−𝜋6+𝜋3 =𝝅𝟔

Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5

Misc 6

Misc 7

Misc 8 Important

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21

Misc 22

Misc 23

Misc 24 Important

Misc 25

Misc 26

Misc 27 You are here

Misc 28

Misc 29

Misc 30 Important

Misc 31

Misc 32 Important

Misc 33

Misc 34

Misc 35

Misc 36

Misc 37

Misc 38

Misc 39

Misc 40

Misc 41 Important

Misc 42

Misc 43

Misc 44 Important

Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.