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Ex 7.9, 8 - Find Integral pi/6 -> pi/4 cosec x - Teachoo

Ex 7.9, 8 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 8 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.9, 8 - Chapter 7 Class 12 Integrals - Part 4

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Ex 7.9, 8 ∫_(πœ‹/6)^(πœ‹/4)β–’γ€–π‘π‘œπ‘ π‘’π‘ π‘₯〗⁑𝑑π‘₯ Let F(π‘₯)=∫1β–’γ€–π‘π‘œπ‘ π‘’π‘ π‘₯ . 𝑑π‘₯γ€— Multiplying and Dividing by π‘π‘œπ‘ π‘’π‘ π‘₯+π‘π‘œπ‘‘ π‘₯ F(π‘₯)=∫1β–’(π‘π‘œπ‘ π‘’π‘ π‘₯ (π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯))/(π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯) 𝑑π‘₯ Let cπ‘œπ‘ π‘’π‘ π‘₯+π‘π‘œπ‘‘ π‘₯=𝑑 Differentiating w.r.t. π‘₯ 𝑑/𝑑π‘₯ (π‘π‘œπ‘ π‘’π‘ π‘₯+π‘π‘œπ‘‘ π‘₯)=𝑑𝑑/𝑑π‘₯ βˆ’π‘π‘œπ‘ π‘’π‘^2 π‘₯βˆ’π‘π‘œπ‘ π‘’π‘ π‘₯ π‘π‘œπ‘‘ π‘₯=𝑑𝑑/𝑑π‘₯ βˆ’π‘π‘œπ‘ π‘’π‘ π‘₯(π‘π‘œπ‘ π‘’π‘ π‘₯+π‘π‘œπ‘‘ π‘₯)=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’π‘π‘œπ‘ π‘’π‘ π‘₯ (π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯) ) Therefore, ∫1β–’(π‘π‘œπ‘ π‘’π‘ π‘₯(π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯))/(π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯) 𝑑π‘₯ =∫1β–’(π‘π‘œπ‘ π‘’π‘ π‘₯(π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯))/𝑑. 𝑑𝑑/(βˆ’π‘π‘œπ‘ π‘’π‘ π‘₯(π‘π‘œπ‘ π‘’π‘ π‘₯ + π‘π‘œπ‘‘ π‘₯) ) =βˆ’βˆ«1▒𝑑𝑑/𝑑 =βˆ’log⁑〖 |𝑑|γ€— =βˆ’log⁑〖 |π‘π‘œπ‘ π‘’π‘ π‘₯+π‘π‘œπ‘‘ π‘₯|γ€— Hence, F(π‘₯)=βˆ’log⁑|π‘π‘œπ‘ π‘’π‘ π‘₯+cot⁑π‘₯ | Now, ∫_(πœ‹/6)^(πœ‹/4)β–’γ€–π‘π‘œπ‘ π‘’π‘ π‘₯=𝐹(πœ‹/4)βˆ’πΉ(πœ‹/6) γ€— =βˆ’π‘™π‘œπ‘”|π‘π‘œπ‘ π‘’π‘(πœ‹/4)+π‘π‘œπ‘‘(πœ‹/4)|βˆ’ (βˆ’log⁑|π‘π‘œπ‘ π‘’π‘(πœ‹/6)+ π‘π‘œπ‘‘(πœ‹/6)| ) =βˆ’π‘™π‘œπ‘”|√2+1|+π‘™π‘œπ‘”|2+√3| =βˆ’π‘™π‘œπ‘”|2+√3|βˆ’ π‘™π‘œπ‘”|√2+1| (Put 𝑑=π‘π‘œπ‘ π‘’π‘ π‘₯+π‘π‘œπ‘‘ π‘₯) =π‘™π‘œπ‘”|(2 + √3)/(√2 + 1)| =π‘™π‘œπ‘”|(2 + √3)/(√2 + 1)Γ—(2 βˆ’ √3)/(2 βˆ’ √3)| =π‘™π‘œπ‘”[((2)^2 βˆ’ (√3)^2)/((√2 + 1) Γ— (2 βˆ’ √3) )] =π‘™π‘œπ‘”[(4 βˆ’ 3)/(√2 + 1)(2 βˆ’ √3) ] =π‘™π‘œπ‘”[(1 Γ— (√2 βˆ’ 1))/((2 βˆ’ √3)(√2 + 1) Γ—(√2 βˆ’ 1) )] =π‘™π‘œπ‘”[(√2 βˆ’ 1)/(2 βˆ’ √3)[(√2)^2βˆ’ 1^2 ] ] =π’π’π’ˆ[(√𝟐 βˆ’ 𝟏)/(𝟐 βˆ’ βˆšπŸ‘)] (log a βˆ’ log b = log π‘Ž/𝑏)

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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.