Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 6 Find the following integrals: (i) ∫1β–’γ€–sin^3⁑π‘₯ cos^2⁑π‘₯ γ€— 𝑑π‘₯ ∫1β–’γ€–sin^3⁑π‘₯ cos^2⁑π‘₯ γ€— 𝑑π‘₯ Let cos π‘₯=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. βˆ’sin⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=(βˆ’π‘‘π‘‘)/sin⁑π‘₯ Now are equation becomes ∫1β–’γ€–sin^3⁑π‘₯ cos^2⁑π‘₯ γ€— 𝑑π‘₯ Putting value of π‘π‘œπ‘ β‘π‘₯ and 𝑑π‘₯ = ∫1β–’sin^3⁑π‘₯ .𝑑^2. 𝑑π‘₯ = ∫1β–’sin^3⁑π‘₯ .𝑑^2. 𝑑𝑑/(βˆ’sin⁑π‘₯ ) = ∫1β–’sin^3⁑π‘₯/(βˆ’sin⁑π‘₯ ) 𝑑^2. 𝑑𝑑 = β€“βˆ«1β–’sin^2⁑π‘₯ 𝑑^2. 𝑑𝑑 = – ∫1β–’(1βˆ’cos^2⁑π‘₯ ) 𝑑^2. 𝑑𝑑 = – ∫1β–’(1βˆ’π‘‘^2 ) 𝑑^2. 𝑑𝑑 = – ∫1β–’(𝑑^2βˆ’π‘‘^4 ) 𝑑𝑑 = ∫1β–’(βˆ’π‘‘^2+𝑑^4 ) 𝑑𝑑 = ∫1β–’γ€–βˆ’π‘‘^2 γ€—. 𝑑𝑑 + ∫1▒𝑑^4 . 𝑑𝑑 (∴ sin^2⁑π‘₯=1βˆ’cos^2⁑π‘₯) = (γ€–βˆ’π‘‘γ€—^2+1)/(2 + 1)+𝑑^(4 + 1)/(4 + 1)+𝐢 = (βˆ’π‘‘^3)/3 +𝑑^5/5 +𝐢 Putting back value of t = cos x = (βˆ’πŸ)/πŸ‘ 〖𝒄𝒐𝒔〗^πŸ‘β‘π’™ +𝟏/πŸ“ 〖𝒄𝒐𝒔〗^πŸ“β‘π’™ +π‘ͺ 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) Example 6 Find the following integrals (iii) ∫1β–’1/(1 + tan⁑π‘₯ ) 𝑑π‘₯ The given function cannot be integrated by direct substitution, so we have to simplify the given function . Simplifying the given function and integrating. ∫1β–’1/(1 + tan⁑π‘₯ ) .𝑑π‘₯ = ∫1β–’1/(1 + sin⁑π‘₯/cos⁑π‘₯ ) .𝑑π‘₯ = ∫1β–’1/((cos⁑π‘₯ + sin⁑π‘₯)/cos⁑π‘₯ ) .𝑑π‘₯ (π‘ˆπ‘ π‘–π‘›π‘” tan⁑π‘₯=sin⁑π‘₯/cos⁑π‘₯ ) = ∫1β–’cos⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ ) .𝑑π‘₯ Multiplying and dividing by 2 = ∫1β–’(2 cos⁑π‘₯)/(2 (sin⁑π‘₯ + cos⁑π‘₯ ) ) .𝑑π‘₯ = ∫1β–’(cos⁑π‘₯ + cos⁑π‘₯)/(2 (sin⁑π‘₯ + cos⁑π‘₯ ) ) .𝑑π‘₯ Adding and subtracting sin⁑π‘₯ in the numerator = ∫1β–’(cos⁑π‘₯ + cos⁑π‘₯ + sin⁑π‘₯ βˆ’ sin⁑π‘₯)/(2 (sin⁑π‘₯ + cos⁑π‘₯ ) ) .𝑑π‘₯ = 1/2 ∫1β–’(sin⁑π‘₯ + cos⁑π‘₯ + cos⁑π‘₯ βˆ’ sin⁑π‘₯)/(sin⁑π‘₯ + cos⁑π‘₯ ) .𝑑π‘₯ = 1/2 ∫1β–’[(sin⁑π‘₯ + cos⁑π‘₯)/(sin⁑π‘₯ + cos⁑π‘₯ )+γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ )] 𝑑π‘₯ = 1/2 ∫1β–’[1+γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ )] 𝑑π‘₯ = 1/2 [∫1β–’γ€–1.𝑑π‘₯+∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ )γ€— 𝑑π‘₯] = 1/2 [π‘₯+𝐢1+∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ ) 𝑑π‘₯] Take, I1 =∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ ) .𝑑π‘₯ Let sin⁑π‘₯ + cos⁑π‘₯=𝑑 Differentiate both sides 𝑀.π‘Ÿ.𝑑.π‘₯. cos⁑π‘₯βˆ’sin⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(cos⁑π‘₯ βˆ’ sin⁑π‘₯ ) Putting value of (𝑠𝑖𝑛⁑π‘₯+π‘π‘œπ‘ β‘π‘₯ ) and 𝑑π‘₯ in I1 . I1=∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ ) .𝑑π‘₯ I1 = ∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/𝑑 .𝑑𝑑/(cos⁑π‘₯ βˆ’ sin⁑π‘₯ ) I1 =∫1β–’1/𝑑 .𝑑𝑑 Let sin⁑π‘₯ + cos⁑π‘₯=𝑑 Differentiate both sides 𝑀.π‘Ÿ.𝑑.π‘₯. cos⁑π‘₯βˆ’sin⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(cos⁑π‘₯ βˆ’ sin⁑π‘₯ ) Putting value of (𝑠𝑖𝑛⁑π‘₯+π‘π‘œπ‘ β‘π‘₯ ) and 𝑑π‘₯ in I1 . I1=∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/(sin⁑π‘₯ + cos⁑π‘₯ ) .𝑑π‘₯ I1 = ∫1β–’γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/𝑑 .𝑑𝑑/(cos⁑π‘₯ βˆ’ sin⁑π‘₯ ) I1 =∫1β–’1/𝑑 .𝑑𝑑 (π‘ˆπ‘ π‘–π‘›π‘” ∫1β–’γ€–1/π‘₯ .γ€— 𝑑π‘₯=log⁑〖 |π‘₯|γ€—+𝐢) (π‘ˆπ‘ π‘–π‘›π‘” 𝑑=sin⁑π‘₯+cos⁑π‘₯ ) (π‘Šβ„Žπ‘’π‘Ÿπ‘’ 𝐢=𝐢1/2+𝐢2/2)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.