Slide1.JPG

Slide2.JPG
Slide3.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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)

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.