Ex 7.3, 6 - Chapter 7 Class 12 Integrals (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 7 Class 12 Integrals
Ex 7.1, 18 Important
Ex 7.1, 20
Ex 7.2, 20 Important
Ex 7.2, 26 Important
Ex 7.2, 35
Ex 7.2, 36 Important
Ex 7.3, 6 Important You are here
Ex 7.3, 13 Important
Ex 7.3, 18 Important
Ex 7.3, 22 Important
Ex 7.3, 24 (MCQ) Important
Example 9 (i)
Example 10 (i)
Ex 7.4, 8 Important
Ex 7.4, 15 Important
Ex 7.4, 21 Important
Ex 7.4, 22
Ex 7.4, 25 (MCQ) Important
Example 15 Important
Ex 7.5, 9 Important
Ex 7.5, 11 Important
Ex 7.5, 17
Ex 7.5, 18 Important
Ex 7.5, 21 Important
Example 20 Important
Example 22 Important
Ex 7.6, 13 Important
Ex 7.6, 14 Important
Ex 7.6, 18 Important
Ex 7.6, 19
Ex 7.6, 24 (MCQ) Important
Ex 7.7, 5 Important
Ex 7.7, 10
Ex 7.7, 11 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Example 25 (i)
Ex 7.8, 15
Ex 7.8, 16 Important
Ex 7.8, 20 Important
Ex 7.8, 22 (MCQ)
Ex 7.9, 4
Ex 7.9, 7 Important
Ex 7.9, 8
Ex 7.9, 9 (MCQ) Important
Example 28 Important
Example 32 Important
Example 34 Important
Ex 7.10,8 Important
Ex 7.10, 18 Important
Example 38 Important
Example 39 Important
Example 42 Important
Misc 18 Important
Misc 8 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Misc 23 Important
Misc 29 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Misc 38 (MCQ) Important
Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams
Integration Formula Sheet - Chapter 41 Class 41 Formulas Important
Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Ex 7.3, 6 𝑠𝑖𝑛 𝑥 sin2𝑥 sin3𝑥 ∫1▒sin〖𝑥 sin〖2𝑥 sin3𝑥 〗 〗 𝑑𝑥 =∫1▒〖(sin𝑥 sin2𝑥 ) sin3𝑥 〗 𝑑𝑥 We know that 2 sin𝐴 sin𝐵=−cos(𝐴+𝐵)+cos(𝐴−𝐵) sin𝐴 sin𝐵=1/2 [−cos(𝐴+𝐵)+cos(𝐴−𝐵) ] sin𝐴 sin𝐵=1/2 [cos(𝐴−𝐵)−cos(𝐴+𝐵) ] Replace A by 𝑥 & B by 2𝑥 sin𝑥 sin2𝑥=1/2 [cos(𝑥−2𝑥)−cos(𝑥+2𝑥) ] sin𝑥 sin2𝑥 =1/2 [cos(−𝑥)−cos(3𝑥) ] sin𝑥 sin2𝑥 =1/2 [cos〖 𝑥〗−cos3𝑥 ] Thus, our equation becomes ∫1▒𝐬𝐢𝐧〖𝒙 𝐬𝐢𝐧𝟐𝒙 sin3𝑥 〗 𝑑𝑥 =∫1▒〖𝟏/𝟐 (𝒄𝒐𝒔𝒙−𝒄𝒐𝒔𝟑𝒙 ) 〗 . sin3𝑥.𝑑𝑥 =1/2 ∫1▒(cos𝑥−cos3𝑥 ) sin3𝑥 𝑑𝑥 =1/2 [∫1▒(cos𝑥. sin3𝑥−cos3𝑥. sin3𝑥 ) ]𝑑𝑥 =1/2 [∫1▒〖cos𝑥. sin3𝑥 〗 𝑑𝑥−∫1▒〖cos3𝑥. sin3𝑥 〗 𝑑𝑥] (∵𝑐𝑜𝑠(−𝑥)=𝑐𝑜𝑠𝑥) ∫1▒〖𝒄𝒐𝒔𝒙. 𝒔𝒊𝒏𝟑𝒙 〗 𝒅𝒙 We know that 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 =𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵=1/2 [𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) ] Replace A by 3𝑥 & B by 𝑥 sin3𝑥 cos𝑥 = 1/2 [𝑠𝑖𝑛(𝑥+3𝑥)+sin(3𝑥−𝑥) ] = 1/2 [𝑠𝑖𝑛4𝑥+sin2𝑥 ] ∫1▒〖𝒄𝒐𝒔𝟑𝒙. 𝒔𝒊𝒏𝟑𝒙 〗 𝒅𝒙 We know that 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 =𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 =1/2 [𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) ] Replace A by 3𝑥 & B by 3𝑥 sin3𝑥 cos3𝑥 = 1/2 [𝑠𝑖𝑛(3𝑥+3𝑥)+sin(3𝑥−3𝑥) ] = 1/2 [𝑠𝑖𝑛6𝑥+sin0 ] =1/2 [𝑠𝑖𝑛6𝑥 ] Hence ∫1▒〖sin3𝑥.cos𝑥 〗 𝑑𝑥 =1/2 ∫1▒[𝑠𝑖𝑛4𝑥+sin2𝑥 ] 𝑑𝑥 Hence ∫1▒〖cos3𝑥.sin3𝑥 〗 𝑑𝑥 =1/2 ∫1▒sin6𝑥 𝑑𝑥 Thus, our equation becomes ∫1▒sin〖𝑥 sin〖2𝑥 sin3𝑥 〗 〗 𝑑𝑥 =1/2 [∫1▒〖sin3𝑥 cos3𝑥 〗 𝑑𝑥−∫1▒〖sin3𝑥 cos3𝑥 〗 𝑑𝑥] =1/2 [1/2 ∫1▒(sin4𝑥+sin2𝑥 ) 𝑑𝑥−1/2 ∫1▒(sin6𝑥 ) 𝑑𝑥] =1/4 [∫1▒(sin4𝑥+sin2𝑥 ) 𝑑𝑥−∫1▒(sin6𝑥 ) 𝑑𝑥] =1/4 [∫1▒sin4𝑥 𝑑𝑥+∫1▒sin2𝑥 𝑑𝑥−∫1▒sin6𝑥 𝑑𝑥] ∫1▒sin(𝑎𝑥+𝑏) 𝑑𝑥=−𝑐𝑜𝑠(𝑎𝑥 + 𝑏)/𝑎 +𝐶 =1/4 [(−cos4𝑥)/4 +(〖−cos〗2𝑥/2) −((−cos6𝑥)/6)]+𝐶 =1/4 [(−cos4𝑥)/4 −cos2𝑥/2+cos6𝑥/6]+𝐶 =𝟏/𝟒 [𝒄𝒐𝒔𝟔𝒙/𝟔 −𝒄𝒐𝒔𝟒𝒙/𝟒 − 𝒄𝒐𝒔𝟐𝒙/𝟐 ]+𝑪