Example 6 - Chapter 7 Class 12 Integrals - Part 4

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Example 6 - Chapter 7 Class 12 Integrals - Part 5

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Example 6 - Chapter 7 Class 12 Integrals - Part 6 Example 6 - Chapter 7 Class 12 Integrals - Part 7

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Example 6 Find the following integrals (ii) ∫1β–’sin⁑π‘₯/sin⁑(π‘₯ + π‘Ž) 𝑑π‘₯ Let π‘₯+π‘Ž=𝑑 Differentiate both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 1=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑 Hence, our equation becomes ∫1β–’sin⁑π‘₯/sin⁑(π‘₯ + π‘Ž) 𝑑π‘₯ Putting the value of (π‘₯⁑+ π‘Ž) and 𝑑π‘₯ = ∫1β–’sin⁑π‘₯/sin⁑𝑑 𝑑π‘₯" " = ∫1β–’π’”π’Šπ’β‘(𝒕 βˆ’ 𝒂)/sin⁑𝑑 𝑑𝑑" " = ∫1β–’(π’”π’Šπ’β‘π’• πœπ¨π¬β‘π’‚ βˆ’ 𝐬𝐒𝐧⁑𝒂 πœπ¨π¬β‘π’•)/sin⁑𝑑 𝑑𝑑 = ∫1β–’(sin⁑〖𝑑 cosβ‘π‘Ž γ€—/sin⁑𝑑 βˆ’ (sinβ‘π‘Ž cos⁑𝑑)/sin⁑𝑑 ) 𝑑𝑑 = ∫1β–’sin⁑〖𝑑 cosβ‘π‘Ž γ€—/sin⁑𝑑 . π‘‘π‘‘βˆ’βˆ«1β–’(sinβ‘π‘Ž cos⁑𝑑)/sin⁑𝑑 . 𝑑𝑑 = ∫1β–’cosβ‘π‘Ž . π‘‘π‘‘βˆ’βˆ«1β–’γ€–sinβ‘π‘Ž co𝑑⁑𝑑 γ€— . 𝑑𝑑 = cos π‘Žβˆ«1β–’1. π‘‘π‘‘βˆ’sinβ‘π‘Ž ∫1▒𝒄𝒐𝒕⁑𝒕 . 𝒅𝒕 = cos π‘Ž (𝑑)βˆ’sinβ‘π‘Ž π’π’π’ˆβ‘|π’”π’Šπ’β‘π’• |+ 𝐢 = 𝑑.cosβ‘π‘Žβˆ’π‘ π‘–π‘›β‘π‘Ž π‘™π‘œπ‘”β‘|𝑠𝑖𝑛⁑𝑑 |+ 𝐢 Putting back value of t = x + a (Using x + a = t ∴ x = t – a) (β–ˆ(π‘ˆπ‘ π‘–π‘›π‘” sin⁑(π‘Žβˆ’π‘)=@sinβ‘π‘Ž cosβ‘π‘βˆ’sin⁑𝑏 cosβ‘π‘Ž )) ( ∫1β–’co𝑑⁑π‘₯ 𝑑π‘₯=log⁑|sin⁑π‘₯ | ) = ∫1β–’π’”π’Šπ’β‘(𝒕 βˆ’ 𝒂)/sin⁑𝑑 𝑑𝑑" " = ∫1β–’π’”π’Šπ’β‘γ€–π’• πœπ¨π¬β‘π’‚ βˆ’π¬π’π§β‘π’‚ πœπ¨π¬β‘π’• γ€—/sin⁑𝑑 𝑑𝑑 = ∫1β–’(sin⁑〖𝑑 cosβ‘π‘Ž γ€—/sin⁑𝑑 βˆ’ (sinβ‘π‘Ž cos⁑𝑑)/sin⁑𝑑 ) 𝑑𝑑 = ∫1β–’sin⁑〖𝑑 cosβ‘π‘Ž γ€—/sin⁑𝑑 . π‘‘π‘‘βˆ’βˆ«1β–’(sinβ‘π‘Ž cos⁑𝑑)/sin⁑𝑑 . 𝑑𝑑 = ∫1β–’cosβ‘π‘Ž . π‘‘π‘‘βˆ’βˆ«1β–’γ€–sinβ‘π‘Ž co𝑑⁑𝑑 γ€— . 𝑑𝑑 = cos π‘Žβˆ«1β–’1. π‘‘π‘‘βˆ’sinβ‘π‘Ž ∫1▒𝒄𝒐𝒕⁑𝒕 . 𝒅𝒕 = cos π‘Ž (𝑑)βˆ’sinβ‘π‘Ž [π’π’π’ˆβ‘|π’”π’Šπ’β‘π’• | ]+𝐢1 = 𝑑.cosβ‘π‘Žβˆ’sinβ‘π‘Ž [log⁑|sin⁑𝑑 | ]+ 𝐢1 Putting back value of t = x + a = (π‘₯+π‘Ž) cosβ‘π‘Žβˆ’sinβ‘π‘Ž [log⁑|sin⁑(π‘₯+π‘Ž) | ]+ 𝐢1 (Using x + a = t x = t – a) (β–ˆ(π‘ˆπ‘ π‘–π‘›π‘” sin⁑(π‘Žβˆ’π‘)=@sinβ‘π‘Ž cosβ‘π‘βˆ’sin⁑𝑏 cosβ‘π‘Ž )) ( ∫1β–’co𝑑⁑π‘₯ 𝑑π‘₯=log⁑|sin⁑π‘₯ | ) = (π‘₯+π‘Ž) cosβ‘π‘Žβˆ’sinβ‘π‘Ž log⁑〖 |sin⁑(π‘₯+π‘Ž) |γ€—+ 𝐢 = π‘₯ cosβ‘π‘Ž+γ€–π‘Ž cosγ€—β‘π‘Žβˆ’sinβ‘π‘Ž log⁑|sin⁑(π‘₯+π‘Ž) |+𝐢 = π‘₯ cosβ‘π‘Žβˆ’sinβ‘π‘Ž log⁑|sin⁑(π‘₯+π‘Ž) |+〖𝒂 𝒄𝒐𝒔〗⁑𝒂+π‘ͺ = 𝒙 π’„π’π’”β‘π’‚βˆ’π¬π’π§β‘γ€– 𝒂〗 π’π’π’ˆβ‘|π’”π’Šπ’β‘(𝒙+𝒂) |+π‚πŸ (π‘Šβ„Žπ‘’π‘Ÿπ‘’ 𝐢1=π‘Ž cosβ‘π‘Ž+𝐢)

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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.