Check sibling questions

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

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)

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.