Misc 32 - Definite integral x tan x / sec x + tanx - Miscellaneous

Misc 32 - Chapter 7 Class 12 Integrals - Part 2
Misc 32 - Chapter 7 Class 12 Integrals - Part 3
Misc 32 - Chapter 7 Class 12 Integrals - Part 4
Misc 32 - Chapter 7 Class 12 Integrals - Part 5


Transcript

Question 2 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 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.