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Misc 7 Integrate the function sin⁑π‘₯/sin⁑(π‘₯ βˆ’ π‘Ž) Let I = ∫1β–’sin⁑π‘₯/sin⁑(π‘₯ βˆ’ π‘Ž) 𝑑π‘₯ Put t = π‘₯ βˆ’ π‘Ž Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑑/𝑑π‘₯ = 𝑑(π‘₯ βˆ’ π‘Ž)/𝑑π‘₯ 𝑑𝑑/𝑑π‘₯ = 1 𝑑π‘₯ = 𝑑𝑑 Therefore ∫1β–’γ€–sin 〗⁑(𝑑 + π‘Ž)/sin⁑𝑑 𝑑𝑑 = ∫1β–’(sin⁑𝑑 cosβ‘π‘Ž + cos⁑𝑑 sinβ‘π‘Ž)/sin⁑𝑑 𝑑𝑑 = ∫1β–’((sin⁑𝑑 cosβ‘π‘Ž)/sin⁑𝑑 + (cos⁑𝑑 sinβ‘π‘Ž)/sin⁑𝑑 ) 𝑑𝑑 = ∫1β–’cosβ‘π‘Ž 𝑑𝑑 + ∫1β–’π‘π‘œπ‘‘β‘π‘‘ sinβ‘π‘Ž 𝑑𝑑 = cosβ‘π‘Ž ∫1▒𝑑𝑑 + sinβ‘π‘Ž ∫1β–’cot⁑𝑑 𝑑𝑑 = cos a Γ— t + sin a log |sin⁑𝑑 | + C Putting back t = x – a = (x – a) cos a + sin a |sin⁑〖(π‘₯βˆ’π‘Ž)γ€— | + C sin⁑(𝐴+𝐡) =sin⁑𝐴 cos⁑𝐡+cos⁑𝐴 sin⁑𝐡 (𝑆𝑖𝑛𝑐𝑒 π‘ π‘–π‘›β‘π‘Ž,π‘π‘œπ‘ β‘π‘Ž π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘ ) = sin a log |sin⁑〖(π‘₯βˆ’π‘Ž)γ€— | + x cos a βˆ’ a cos a + C = sin a log |𝐬𝐒𝐧⁑〖(π’™βˆ’π’‚)γ€— | + x cos a + 𝐂_𝟏 (𝐢_1= βˆ’ a cos a + C)

  1. Chapter 7 Class 12 Integrals
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About the Author

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 and Computer Science at Teachoo