Chapter 7 Class 12 Integrals
Concept wise

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.