Example 6 - Chapter 7 Class 12 Integrals - Part 8

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Example 6 - Chapter 7 Class 12 Integrals - Part 9

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Example 6 - Chapter 7 Class 12 Integrals - Part 10 Example 6 - Chapter 7 Class 12 Integrals - Part 11 Example 6 - Chapter 7 Class 12 Integrals - Part 12

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

Transcript

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 10 years. He provides courses for Maths and Science at Teachoo.