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 You are here

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

Ex 7.6, 18 - Integrate e^x (1 + sin x / 1 + cos x) - Teachoo

Ex 7.6, 18 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 18 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.6, 18 - Chapter 7 Class 12 Integrals - Part 4


Transcript

Ex 7.6, 18 Integrate the function - 𝑒π‘₯ ((1 + sin⁑π‘₯)/(1 + cos⁑π‘₯ )) Simplifying function 𝑒^π‘₯ ((1 + sin⁑π‘₯)/(1 + cos⁑π‘₯ )) 𝑒^π‘₯ ((1 + sin⁑π‘₯)/(1 + cos⁑π‘₯ ))=𝑒^π‘₯ ((1 + 2 sin⁑(π‘₯/2) cos⁑(π‘₯/2))/(2 γ€–π‘π‘œπ‘ ^2〗⁑(π‘₯/2) )) π’”π’Šπ’β‘πŸπ’™=𝟐 π’”π’Šπ’β‘π’™ 𝒄𝒐𝒔⁑𝒙 Replacing x by π‘₯/2 , we get 〖𝑠𝑖𝑛 2〗⁑(π‘₯/2)=2 𝑠𝑖𝑛⁑(π‘₯/2) π‘π‘œπ‘ β‘(π‘₯/2) 𝑠𝑖𝑛⁑π‘₯=2 𝑠𝑖𝑛⁑(π‘₯/2) π‘π‘œπ‘ β‘(π‘₯/2) πœπ¨π¬β‘πŸπ’™= 2γ€–"cos" γ€—^𝟐 πœ½βˆ’πŸ Replacing πœƒπ‘₯ by π‘₯/2 γ€–π‘π‘œπ‘  2〗⁑(π‘₯/2)=2π‘π‘œπ‘ ^2 (π‘₯/2)βˆ’1 1+π‘π‘œπ‘ β‘π‘₯=2π‘π‘œπ‘ ^2 π‘₯/2 We know 〖𝑠𝑖𝑛^2〗⁑π‘₯+γ€–π‘π‘œπ‘ ^2〗⁑π‘₯⁑= 1 Replacing x by π‘₯/2 , we get 〖𝑠𝑖𝑛^2〗⁑(π‘₯/2)⁑〖+γ€–π‘π‘œπ‘ ^2〗⁑(π‘₯/2) γ€—=1 =𝑒^π‘₯ ((γ€–π’”π’Šπ’γ€—^𝟐⁑(𝒙/𝟐) + 〖𝒄𝒐𝒔〗^𝟐⁑(𝒙/𝟐) +2 γ€–sin 〗⁑〖π‘₯/2γ€— . γ€–cos 〗⁑〖π‘₯/2γ€—)/( 2 γ€–π‘π‘œπ‘ ^2〗⁑(π‘₯/2) )) =𝑒^π‘₯ ((γ€–sin 〗⁑〖π‘₯/2γ€— + γ€–cos 〗⁑〖π‘₯/2γ€— )^2/( 2 γ€–π‘π‘œπ‘ ^2〗⁑(π‘₯/2) )) =𝑒^π‘₯/2 ((γ€–sin 〗⁑〖π‘₯/2γ€— + γ€–cos 〗⁑〖π‘₯/2γ€—)/( π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— ))^2 =𝑒^π‘₯/2 (γ€–sin 〗⁑〖π‘₯/2γ€—/( π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— ) + γ€–cos 〗⁑〖π‘₯/2γ€—/( π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— ))^2 =𝑒^π‘₯/2 (γ€–tan 〗⁑〖π‘₯/2γ€— +1)^2 =𝑒^π‘₯/2 (tan^2⁑〖π‘₯/2γ€— +(1)^2+2(γ€–tan 〗⁑〖π‘₯/2γ€— )(1)) =𝑒^π‘₯/2 (〖〖𝒕𝒂𝒏〗^𝟐 〗⁑〖𝒙/πŸγ€— +𝟏+2 γ€–tan 〗⁑〖π‘₯/2γ€— ) =𝑒^π‘₯/2 (〖〖𝒔𝒆𝒄〗^𝟐 〗⁑〖𝒙/πŸγ€— +2 γ€–tan 〗⁑(π‘₯/2) ) =𝑒^π‘₯ (1/2 . sec^2⁑〖π‘₯/2γ€— +2/2 γ€–tan 〗⁑(π‘₯/2) ) =𝑒^π‘₯ (γ€–tan 〗⁑(π‘₯/2)+1/2 sec^2⁑(π‘₯/2) ) We know γ€–1+tan^2〗⁑π‘₯=sec^2⁑π‘₯ Replacing x by π‘₯/2 , we get γ€–1+γ€–π‘‘π‘Žπ‘›^2〗⁑(π‘₯/2)〗⁑= sec^2 (π‘₯/2) Thus, Our Integration becomes ∫1▒〖𝑒^π‘₯ " " ((1 + sin⁑π‘₯)/(1 + cos⁑π‘₯ )) 𝑑π‘₯γ€— " "=∫1▒〖𝑒^π‘₯ (γ€–tan 〗⁑(π‘₯/2)+1/2 sec^2⁑(π‘₯/2) )𝑑π‘₯γ€— " " It is of the form ∫1▒〖𝑒^π‘₯ [𝑓(π‘₯)+𝑓^β€² (π‘₯)] γ€— 𝑑π‘₯=𝑒^π‘₯ 𝑓(π‘₯)+𝐢 Where 𝑓(π‘₯)=tan^(βˆ’1)⁑π‘₯ 𝑓^β€² (π‘₯)= 1/(1 + π‘₯^2 ) So, our equation becomes ∫1▒〖𝑒^π‘₯ " " ((1 + sin⁑π‘₯)/(1 + cos⁑π‘₯ )) 𝑑π‘₯γ€— " " = 𝒆^𝒙 〖𝒕𝒂𝒏 〗⁑〖𝒙/πŸγ€—+π‘ͺ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.