Chapter 7 Class 12 Integrals

Ex 7.1, 10 Important

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

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Transcript

Misc 18 Integrate the function 1/√(sin^3⁡𝑥 sin⁡(𝑥 + 𝛼) ) Solving sin^3⁡𝑥 sin⁡(𝑥 + 𝛼) =sin^3⁡𝑥 [sin⁡𝑥 cos⁡𝛼+cos⁡𝑥.sin⁡𝛼 ] =〖sin^4 𝑥〗⁡cos⁡𝛼 +cos⁡𝑥.sin^3⁡𝑥 sin⁡𝛼 =〖sin^4 𝑥〗⁡cos⁡𝛼 +cos⁡𝑥.sin^3⁡𝑥 sin⁡𝛼×sin⁡𝑥/sin⁡𝑥 =sin^4⁡𝑥 [cos⁡𝛼+cos⁡𝑥 . sin⁡𝛼.1/sin⁡𝑥 ] Hence Using 𝑠𝑖𝑛⁡(𝐴+𝐵)=𝑠𝑖𝑛⁡𝐴 𝑐𝑜𝑠⁡𝐵+𝑐𝑜𝑠⁡𝐴.𝑠𝑖𝑛⁡𝐵 =sin^4⁡𝑥 [cos⁡𝛼+cos⁡𝑥/sin⁡𝑥 . sin⁡𝛼 ] =sin^4⁡𝑥 [cos⁡𝛼+cot⁡𝑥 sin⁡𝛼 ] Therefore sin^3⁡𝑥 sin⁡(𝑥+𝛼)=sin^4⁡𝑥 (cos⁡𝛼+cot⁡𝑥.sin⁡𝛼 ) Now ∫1▒1/√(sin^3⁡𝑥 sin⁡(𝑥 + 𝛼) ) 𝑑𝑥 =∫1▒1/√(sin^4⁡𝑥 (cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 ) ) 𝑑𝑥 =∫1▒〖1/√(sin^4⁡𝑥 )×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2⁡𝑥 ×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 Let cos⁡𝛼+cot⁡𝑥. sin⁡𝛼=𝑡 Diff w.r.t. x 𝑑(cos⁡𝛼 + cot⁡𝑥 sin⁡𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos⁡𝛼 )/𝑑𝑥+sin⁡𝛼 𝑑(cot⁡𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 =∫1▒〖1/√(sin^4⁡𝑥 )×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 =∫1▒〖1/sin^2⁡𝑥 ×1/√(cos⁡𝛼 + cot⁡𝑥 . sin⁡𝛼 )〗 𝑑𝑥 Let cos⁡𝛼+cot⁡𝑥. sin⁡𝛼=𝑡 Diff w.r.t. x 𝑑(cos⁡𝛼 + cot⁡𝑥 sin⁡𝛼 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 𝑑(cos⁡𝛼 )/𝑑𝑥+sin⁡𝛼 𝑑(cot⁡𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 0+sin⁡𝛼 (−𝑐𝑜𝑠𝑒𝑐^2 𝑥)=𝑑𝑡/𝑑𝑥 −sin⁡𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(−sin⁡𝛼 𝑐𝑜𝑠𝑒𝑐^2 𝑥) 𝑑𝑥=1/(−sin⁡𝛼 ) . 1/(𝑐𝑜𝑠𝑒𝑐^2 𝑥) . 𝑑𝑡 𝑑𝑥=1/(−sin⁡𝛼 ) . sin^2⁡𝑥. 𝑑𝑡 Now our equation becomes ∫1▒1/(sin^2⁡𝑥 √(cos⁡𝛼 + cot⁡𝑥 sin⁡𝛼 ) ) 𝑑𝑥 =∫1▒1/(sin^2⁡𝑥 √𝑡 )×1/(−sin⁡𝛼 )×sin^2⁡𝑥 𝑑𝑡 cos⁡𝛼 &sin⁡𝛼 "are constant" █(" " @"&" 𝑑(cot⁡𝑥 )/𝑑𝑥=−𝑐𝑜𝑠𝑒𝑐 𝑥) =1/(−sin⁡𝛼 ) ∫1▒1/√𝑡 𝑑𝑡 =(−1)/sin⁡𝛼 ∫1▒(𝑡)^((−1)/2) 𝑑𝑡 =(−1)/sin⁡𝛼 [𝑡^((−1)/2 + 1)/((−1)/2 + 1) +𝐶] =(−1)/sin⁡𝛼 [𝑡^(1/2)/(1/2) +𝐶] =(−𝟏)/𝒔𝒊𝒏⁡𝜶 [𝟐√𝒕 +𝑪] Putting back value of 𝑡=√(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) =(−1)/sin⁡𝛼 [2√(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 )+𝐶] =(−2)/sin⁡𝛼 √(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) −1/sin⁡𝛼 . 𝐶 =(−2)/sin⁡𝛼 √(𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) +𝐶 Now, From (1) sin^3⁡𝑥 sin⁡(𝑥+𝛼)=sin^4⁡𝑥 (𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 ) (sin^3⁡𝑥 sin⁡(𝑥 + 𝛼))/sin^4⁡𝑥 =𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 〖𝑠𝑖𝑛 〗⁡(𝑥 + 𝛼)/sin⁡𝑥 =𝑐𝑜𝑠⁡𝛼+𝑐𝑜𝑡⁡𝑥. 𝑠𝑖𝑛⁡𝛼 Thus, Answer =(−𝟐)/𝒔𝒊𝒏⁡𝒙 √(𝒔𝒊𝒏⁡(𝒙 + 𝜶)/𝐬𝐢𝐧⁡𝒙 ) + 𝑪

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.