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


Misc 32 Evaluate the definite integral ∫_0^πœ‹β–’(π‘₯ tan⁑π‘₯ )/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ Let I=∫_0^πœ‹β–’(π‘₯ tan⁑π‘₯ )/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ ∴ I=∫_0^πœ‹β–’((πœ‹ βˆ’ π‘₯) tan⁑〖 (πœ‹ βˆ’ π‘₯)γ€—)/(sec⁑(πœ‹ βˆ’ π‘₯) +γ€– tan〗⁑(πœ‹ βˆ’ π‘₯) ) 𝑑π‘₯ I=∫_0^πœ‹β–’((πœ‹ βˆ’ π‘₯)(βˆ’tan⁑〖 π‘₯γ€—) )/((βˆ’sec⁑〖 π‘₯γ€—) + γ€–( βˆ’tan〗⁑π‘₯)) 𝑑π‘₯ I=∫_0^πœ‹β–’(βˆ’(πœ‹ βˆ’ π‘₯) tan⁑π‘₯ )/(βˆ’(sec⁑π‘₯ +γ€– tan〗⁑π‘₯)) 𝑑π‘₯ Using The Property, P4 P4 : ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=γ€— ∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ I=∫_0^πœ‹β–’((πœ‹ βˆ’ π‘₯) tan⁑π‘₯ )/((sec⁑π‘₯ +γ€– tan〗⁑π‘₯)) 𝑑π‘₯ Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^πœ‹β–’(π‘₯ tan⁑π‘₯ )/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯+∫_0^πœ‹β–’(πœ‹ tan⁑π‘₯ βˆ’ π‘₯ tan⁑π‘₯)/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ 2I=∫_0^πœ‹β–’(π‘₯ tan⁑π‘₯ + πœ‹ tan⁑π‘₯ βˆ’ π‘₯ tan⁑π‘₯)/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ 2I=∫_0^πœ‹β–’(πœ‹ tan⁑π‘₯)/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ 2I=πœ‹βˆ«_0^πœ‹β–’tan⁑π‘₯/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’tan⁑π‘₯/(sec⁑π‘₯ +γ€– tan〗⁑π‘₯ ) 𝑑π‘₯ =πœ‹/2 ∫_0^πœ‹β–’(sin⁑π‘₯/cos⁑π‘₯ )/(1/cos⁑π‘₯ + sin⁑π‘₯/cos⁑π‘₯ ) 𝑑π‘₯ =πœ‹/2 ∫_0^πœ‹β–’sin⁑π‘₯/(1 + sin⁑π‘₯ ) 𝑑π‘₯ =πœ‹/2 ∫_0^πœ‹β–’(sin⁑π‘₯ + 1 βˆ’ 1)/(1 + sin⁑π‘₯ ) 𝑑π‘₯ =πœ‹/2 ∫_0^πœ‹β–’[(1 + sin⁑π‘₯)/(1 + sin⁑π‘₯ ) βˆ’1/(1 + sin⁑π‘₯ )] 𝑑π‘₯ =πœ‹/2 ∫_0^πœ‹β–’[1 βˆ’1/(1 + sin⁑π‘₯ )] 𝑑π‘₯ =πœ‹/2 [∫_0^πœ‹β–’1 𝑑π‘₯βˆ’βˆ«_0^πœ‹β–’1/(1 + sin⁑π‘₯ ) 𝑑π‘₯] =πœ‹/2 [[π‘₯]_0^πœ‹βˆ’βˆ«_0^πœ‹β–’1/(1 + sin⁑π‘₯ ) ((1 βˆ’ sin⁑π‘₯)/(1 βˆ’ sin⁑π‘₯ )) 𝑑π‘₯] =πœ‹/2 [[πœ‹βˆ’0]βˆ’βˆ«_0^πœ‹β–’(1 βˆ’ sin⁑π‘₯)/(1 βˆ’ sin^2⁑π‘₯ ) 𝑑π‘₯] =πœ‹/2 [πœ‹βˆ’βˆ«_0^πœ‹β–’(1 βˆ’ sin⁑π‘₯)/cos^2⁑π‘₯ 𝑑π‘₯] =πœ‹/2 [πœ‹βˆ’βˆ«_0^πœ‹β–’[1/cos^2⁑π‘₯ βˆ’ sin⁑π‘₯/cos^2⁑π‘₯ ] 𝑑π‘₯] =πœ‹/2 {πœ‹βˆ’βˆ«_0^πœ‹β–’[sec^2⁑π‘₯βˆ’tan⁑π‘₯ sec⁑π‘₯ ] 𝑑π‘₯} =πœ‹/2 {πœ‹βˆ’βˆ«_0^πœ‹β–’sec^2⁑π‘₯ 𝑑π‘₯+∫_0^πœ‹β–’γ€–tan⁑π‘₯ sec⁑π‘₯ γ€— 𝑑π‘₯} =πœ‹/2 [πœ‹βˆ’[tan⁑π‘₯ ]_0^πœ‹+[sec⁑π‘₯ ]_0^πœ‹ ] =πœ‹/2 {πœ‹βˆ’[tan⁑〖(πœ‹)βˆ’tan⁑(0) γ€— ]+[sec (πœ‹)βˆ’sec⁑(0) ]} =πœ‹/2 {πœ‹βˆ’[0βˆ’0]+[βˆ’1βˆ’1]} =πœ‹/2 {πœ‹βˆ’0+[βˆ’2]} =𝝅/𝟐 (π…βˆ’πŸ)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.