Β  Misc 24 - Evaluate definite integral ex (1-sin x / 1-cos x) - Miscellaneous

part 2 - Misc 24 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Misc 24 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals

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Misc 24 Evaluate the definite integral ∫_(πœ‹/2)^πœ‹β–’γ€–e^π‘₯⁑((1 βˆ’sin⁑π‘₯)/(1 βˆ’cos⁑π‘₯ )) 𝑑π‘₯γ€— ∫_(πœ‹/2)^πœ‹β–’γ€–e^π‘₯⁑((1 βˆ’ sin⁑π‘₯)/(1 βˆ’γ€– cos〗⁑π‘₯ )) 𝑑π‘₯γ€— = ∫_(πœ‹/2)^πœ‹β–’γ€–e^π‘₯⁑((1 )/(1 βˆ’ cos⁑π‘₯ )βˆ’sin⁑π‘₯/(1 βˆ’ cos⁑π‘₯ )) 𝑑π‘₯γ€— = ∫_(πœ‹/2)^πœ‹β–’γ€–e^π‘₯⁑((βˆ’sin⁑π‘₯)/(1 βˆ’ cos⁑π‘₯ )+(1 )/(1 βˆ’ cos⁑π‘₯ )) 𝑑π‘₯γ€— Let f(x) = (βˆ’sin⁑π‘₯)/(1 βˆ’ cos⁑π‘₯ ) f’(x) = βˆ’[(cos⁑π‘₯ (1 βˆ’ cos⁑〖π‘₯) βˆ’ (sin⁑〖π‘₯) (sin⁑〖π‘₯)γ€— γ€— γ€—)/((1 βˆ’ γ€–cos⁑π‘₯)γ€—^2 )] =βˆ’[(cos⁑〖π‘₯ βˆ’γ€— γ€–π‘π‘œπ‘ γ€—^2 π‘₯ βˆ’ 〖𝑠𝑖𝑛〗^2 π‘₯)/((1 βˆ’ γ€–cos π‘₯⁑)γ€—^2 )] = βˆ’(cos⁑〖π‘₯ βˆ’ (γ€–π‘π‘œπ‘ γ€—^2 π‘₯ + 〖𝑠𝑖𝑛〗^2 π‘₯)γ€—/((1 βˆ’ γ€–π‘π‘œπ‘  π‘₯⁑)γ€—^2 )) = βˆ’ (cos⁑〖π‘₯ βˆ’ 1γ€—/γ€–(1 βˆ’γ€– cos〗⁑〖π‘₯)γ€—γ€—^2 ) = ( 1 βˆ’ cos⁑〖π‘₯ γ€—)/γ€–(1 βˆ’ cos⁑〖π‘₯)γ€—γ€—^2 = 1/(1 βˆ’ cos⁑π‘₯ ) Hence, the given integration is of form, ∫1▒〖𝑒^π‘₯ (𝑓(π‘₯)+𝑓^β€² (π‘₯)) 𝑑π‘₯γ€—= 𝑒^π‘₯ 𝑓(π‘₯) Where f(x) = (βˆ’sin⁑π‘₯)/(1 βˆ’ cos⁑π‘₯ ) and f’(x) = 1/(1 βˆ’ cos⁑π‘₯ ) Hence, ∫_(πœ‹/2)^πœ‹β–’γ€–e^π‘₯⁑((1 βˆ’ sin⁑π‘₯)/(1 βˆ’γ€– cos〗⁑π‘₯ )) 𝑑π‘₯γ€— = ∫_(πœ‹/2)^πœ‹β–’γ€–e^π‘₯⁑((1 )/(1 βˆ’ cos⁑π‘₯ )βˆ’sin⁑π‘₯/(1 βˆ’ cos⁑π‘₯ )) 𝑑π‘₯γ€— = [𝑒^π‘₯ ((βˆ’sin⁑π‘₯)/(1 βˆ’cos⁑π‘₯ ))]_(πœ‹/2)^πœ‹ = ((𝑒^π‘₯ sin⁑π‘₯)/cos⁑〖π‘₯ βˆ’1γ€— )_(πœ‹/2)^πœ‹ Putting limits = (𝑒^πœ‹ sinβ‘πœ‹)/cosβ‘γ€–πœ‹ βˆ’1γ€— βˆ’ (𝑒^(πœ‹/2) sinβ‘γ€–πœ‹/2γ€—)/cos⁑〖 πœ‹/2 βˆ’1γ€— = (𝑒^πœ‹ Γ— (0))/(βˆ’1 βˆ’1)βˆ’ (𝑒^(πœ‹/2) (1))/(0 βˆ’ 1) = 𝒆^(𝝅/𝟐)

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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 and Computer Science at Teachoo