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Misc 25 - Evaluate definite integral ex (1-sin x / 1-cos x)

Misc 25 - Chapter 7 Class 12 Integrals - Part 2
Misc 25 - Chapter 7 Class 12 Integrals - Part 3

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Misc 25 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.