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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
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 You are here
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 May 29, 2023 by Teachoo
Misc 18 Integrate the function 1/√(sin^3𝑥 sin(𝑥 + 𝛼) ) Solving sin^3𝑥 sin(𝑥 + 𝛼) =sin^3𝑥 [sin𝑥 cos𝛼+cos𝑥.sin𝛼 ] =〖sin^4 𝑥〗cos𝛼 +cos𝑥.sin^3𝑥 sin𝛼 =〖sin^4 𝑥〗cos𝛼 +cos𝑥.sin^3𝑥 sin𝛼×sin𝑥/sin𝑥 =sin^4𝑥 [cos𝛼+cos𝑥 . sin𝛼.1/sin𝑥 ] Hence Using 𝑠𝑖𝑛(𝐴+𝐵)=𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵+𝑐𝑜𝑠𝐴.𝑠𝑖𝑛𝐵 =sin^4𝑥 [cos𝛼+cos𝑥/sin𝑥 . sin𝛼 ] =sin^4𝑥 [cos𝛼+cot𝑥 sin𝛼 ] Therefore sin^3𝑥 sin(𝑥+𝛼)=sin^4𝑥 (cos𝛼+cot𝑥.sin𝛼 ) Now ∫1▒1/√(sin^3𝑥 sin(𝑥 + 𝛼) ) 𝑑𝑥 =∫1▒1/√(sin^4𝑥 (cos𝛼 + cot𝑥 . sin𝛼 ) ) 𝑑𝑥 =∫1▒〖1/√(sin^4𝑥 )×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2𝑥 ×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 Let cos𝛼+cot𝑥. sin𝛼=𝑡 Diff w.r.t. x 𝑑(cos𝛼 + cot𝑥 sin𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos𝛼 )/𝑑𝑥+sin𝛼 𝑑(cot𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 =∫1▒〖1/√(sin^4𝑥 )×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2𝑥 ×1/√(cos𝛼 + cot𝑥 . sin𝛼 )〗 𝑑𝑥 Let cos𝛼+cot𝑥. sin𝛼=𝑡 Diff w.r.t. x 𝑑(cos𝛼 + cot𝑥 sin𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos𝛼 )/𝑑𝑥+sin𝛼 𝑑(cot𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 0+sin𝛼 (−𝑐𝑜𝑠𝑒𝑐^2 𝑥)=𝑑𝑡/𝑑𝑥 −sin𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(−sin𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥) 𝑑𝑥=1/(−sin𝛼 ) . 1/(𝑐𝑜𝑠𝑒𝑐^2 𝑥) . 𝑑𝑡 𝑑𝑥=1/(−sin𝛼 ) . sin^2𝑥. 𝑑𝑡 Now our equation becomes ∫1▒1/(sin^2𝑥 √(cos𝛼 + cot𝑥 sin𝛼 ) ) 𝑑𝑥 =∫1▒1/(sin^2𝑥 √𝑡 )×1/(−sin𝛼 )×sin^2𝑥 𝑑𝑡 cos𝛼 &sin𝛼 "are constant" █(" " @"&" 𝑑(cot𝑥 )/𝑑𝑥=−𝑐𝑜𝑠𝑒𝑐 𝑥) =1/(−sin𝛼 ) ∫1▒1/√𝑡 𝑑𝑡 =(−1)/sin𝛼 ∫1▒(𝑡)^((−1)/2) 𝑑𝑡 =(−1)/sin𝛼 [𝑡^((−1)/2 + 1)/((−1)/2 + 1) +𝐶] =(−1)/sin𝛼 [𝑡^(1/2)/(1/2) +𝐶] =(−𝟏)/𝒔𝒊𝒏𝜶 [𝟐√𝒕 +𝑪] Putting back value of 𝑡=√(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) =(−1)/sin𝛼 [2√(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 )+𝐶] =(−2)/sin𝛼 √(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) −1/sin𝛼 . 𝐶 =(−2)/sin𝛼 √(𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) +𝐶 Now, From (1) sin^3𝑥 sin(𝑥+𝛼)=sin^4𝑥 (𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 ) (sin^3𝑥 sin(𝑥 + 𝛼))/sin^4𝑥 =𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 〖𝑠𝑖𝑛 〗(𝑥 + 𝛼)/sin𝑥 =𝑐𝑜𝑠𝛼+𝑐𝑜𝑡𝑥. 𝑠𝑖𝑛𝛼 Thus, Answer =(−𝟐)/𝒔𝒊𝒏𝒙 √(𝒔𝒊𝒏(𝒙 + 𝜶)/𝐬𝐢𝐧𝒙 ) + 𝑪