Example 6 - Chapter 7 Class 12 Integrals - Part 8

Example 6 - Chapter 7 Class 12 Integrals - Part 9
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


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 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.