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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


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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.