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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.