Misc 11 - Integrate 1 / cos (x + a) cos (x + b) - Class 12 - Integration using trigo identities - a-b formulae

Misc 11 - Chapter 7 Class 12 Integrals - Part 2
Misc 11 - Chapter 7 Class 12 Integrals - Part 3
Misc 11 - Chapter 7 Class 12 Integrals - Part 4


Transcript

Misc 11 Integrate the function 1/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) ∫1▒𝑑π‘₯/cos⁑〖(π‘₯ + π‘Ž) cos⁑〖(π‘₯ + 𝑏)γ€— γ€— Divide & Multiplying by 𝐬𝐒𝐧⁑(π’‚βˆ’π’ƒ) =∫1β–’γ€–sin⁑(π‘Ž βˆ’ 𝑏)/sin⁑(π‘Ž βˆ’ 𝑏) Γ— 1/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) )γ€— 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’sin⁑(π‘Ž βˆ’ 𝑏)/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’sin⁑(π‘Ž βˆ’ 𝑏 + π‘₯ βˆ’ π‘₯)/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’(γ€–sin 〗⁑〖((π‘₯ + π‘Ž)γ€— βˆ’ (π‘₯ + 𝑏)) )/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) 𝑑π‘₯ We know that 𝑠𝑖𝑛⁑(π΄βˆ’π΅)=𝑠𝑖𝑛⁑𝐴 π‘π‘œπ‘ β‘π΅βˆ’π‘π‘œπ‘ β‘π΄ 𝑠𝑖𝑛⁑𝐡 Replace A by (π‘₯+π‘Ž) & B by (π‘₯+𝑏) 𝑠𝑖𝑛⁑((π‘₯+π‘Ž)βˆ’(π‘₯+𝑏))=𝑠𝑖𝑛⁑(π‘₯+π‘Ž) π‘π‘œπ‘ β‘(π‘₯+𝑏)βˆ’π‘π‘œπ‘ β‘(π‘₯+π‘Ž) 𝑠𝑖𝑛⁑(π‘₯+𝑏) =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’(𝑠𝑖𝑛⁑(π‘₯ + π‘Ž) π‘π‘œπ‘ β‘(π‘₯ + 𝑏) βˆ’ π‘π‘œπ‘ β‘(π‘₯ + π‘Ž) 𝑠𝑖𝑛⁑(π‘₯ + 𝑏)" " )/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’((𝑠𝑖𝑛⁑(π‘₯ + π‘Ž) π‘π‘œπ‘ β‘(π‘₯ + 𝑏))/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) βˆ’(cos⁑(π‘₯ + π‘Ž) sin⁑(π‘₯ + 𝑏))/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) )) 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’(𝑠𝑖𝑛⁑(π‘₯ + π‘Ž)/cos⁑(π‘₯ + π‘Ž) βˆ’π‘ π‘–π‘›β‘(π‘₯ + 𝑏)/cos⁑(π‘₯ + 𝑏) ) 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’(tan⁑(π‘₯+π‘Ž) βˆ’tan⁑(π‘₯+𝑏) ) 𝑑π‘₯ ∫1▒𝒕𝒂𝒏⁑(𝒙+𝒂) 𝒅𝒙 Let (π‘₯+π‘Ž)=𝑑 Diff w.r.t. x 1+0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑 ∫1β–’tan⁑(π‘₯+π‘Ž) 𝑑π‘₯ =∫1β–’tan⁑𝑑 . 𝑑𝑑 =βˆ’log⁑|cos⁑𝑑 |+𝐢1 Putting value of 𝑑=π‘₯+π‘Ž =βˆ’log⁑|cos⁑〖(π‘₯+π‘Ž)γ€— |+𝐢1 ∫1▒𝒕𝒂𝒏⁑(𝒙+𝒃) 𝒅𝒙 Let (π‘₯+𝑏)=𝑑 Diff w.r.t.x 1+0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑 ∫1β–’tan⁑(π‘₯+𝑏) 𝑑π‘₯ =∫1β–’tan⁑𝑑 . 𝑑𝑑 =βˆ’log⁑|cos⁑𝑑 |+𝐢2 Putting value of 𝑑=π‘₯+𝑏 =βˆ’log⁑|cos⁑〖(π‘₯+𝑏)γ€— |+𝐢2 Thus, our equation becomes ∫1β–’1/(cos⁑(π‘₯ + π‘Ž) cos⁑(π‘₯ + 𝑏) ) 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) ∫1β–’γ€–tan⁑(π‘₯+π‘Ž)βˆ’tan⁑(π‘₯+π‘Ž) γ€— 𝑑π‘₯ =1/sin⁑(π‘Ž βˆ’ 𝑏) [βˆ’log⁑|cos⁑(π‘₯+π‘Ž) |+𝐢1βˆ’(βˆ’log⁑|cos⁑(π‘₯+𝑏) |+𝐢2) =1/sin⁑(π‘Ž βˆ’ 𝑏) [βˆ’log⁑|cos⁑(π‘₯+π‘Ž) |+log⁑|cos⁑(π‘₯+𝑏) |+𝐢1+𝐢2] =1/sin⁑(π‘Ž βˆ’ 𝑏) [βˆ’log⁑|cos⁑(π‘₯+π‘Ž) |+log⁑|cos⁑(π‘₯+𝑏) | ]+𝟏/π’”π’Šπ’β‘(𝒂 + 𝒃) (π‘ͺ𝟏+π‘ͺ𝟐) =1/sin⁑(π‘Ž βˆ’ 𝑏) [βˆ’log⁑|cos⁑(π‘₯+π‘Ž) |+log⁑|cos⁑(π‘₯+𝑏) | ]+π‘ͺ = 𝟏/π’”π’Šπ’β‘(𝒂 βˆ’ 𝒃) π₯𝐨𝐠|𝒄𝒐𝒔⁑(𝒙 + 𝒃)/𝒄𝒐𝒔⁑(𝒙 + 𝒂) |+π‘ͺ ("log a βˆ’ log b = log " π‘Ž/𝑏 " " )

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.