Example 6 - Chapter 7 Class 12 Integrals - Part 4

Example 6 - Chapter 7 Class 12 Integrals - Part 5
Example 6 - Chapter 7 Class 12 Integrals - Part 6 Example 6 - Chapter 7 Class 12 Integrals - Part 7

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 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.