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

Transcript

Ex 7.2, 32 Integrate 1/(1 + cot⁑π‘₯ ) Simplify the given function ∫1β–’1/(1 + cot⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’1/(1 + cos⁑π‘₯/sin⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’1/(γ€–sin⁑π‘₯ + cos〗⁑π‘₯/sin⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’sin⁑π‘₯/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ 𝑑π‘₯ Multiplying & dividing by 2 = ∫1β–’(2 sin⁑π‘₯)/2(γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯ Adding & subtracting π‘π‘œπ‘ β‘π‘₯ in numerator = ∫1β–’(sin⁑π‘₯ + sin⁑π‘₯ + cos⁑π‘₯ βˆ’ cos⁑π‘₯)/2(γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1β–’((sin⁑π‘₯ + cos⁑π‘₯ + sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1β–’((sin⁑π‘₯ + cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ +(sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1β–’(1+(sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 [π‘₯+∫1β–’((sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯] + 𝐢1 …(1) Solving 𝐈1 I1 = ∫1β–’(sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ 𝑑π‘₯ Let γ€–sin⁑π‘₯ + cos〗⁑π‘₯=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ γ€–cos⁑π‘₯βˆ’sin〗⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’(γ€–sin⁑π‘₯ βˆ’ cos〗⁑π‘₯ ) ) Thus, our equation becomes …(2) I1 = ∫1β–’(sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ 𝑑π‘₯ = ∫1β–’(sin⁑π‘₯ βˆ’ cos⁑π‘₯)/𝑑 . 𝑑𝑑/(βˆ’(γ€–sin⁑π‘₯ βˆ’ cos〗⁑π‘₯ ) ) = βˆ’1∫1▒𝑑𝑑/𝑑 = βˆ’γ€–log 〗⁑|𝑑|+𝐢 Putting back 𝑑=𝑠𝑖𝑛⁑π‘₯+π‘π‘œπ‘ β‘π‘₯ = βˆ’log⁑〖 |sin⁑π‘₯+cos⁑π‘₯ |γ€—+𝐢2 Putting the value of I1 in (1) ∴ ∫1β–’γ€–1/(1 + cot⁑π‘₯ ) " " γ€— = 1/2 [π‘₯+∫1β–’((sin⁑π‘₯ βˆ’ cos⁑π‘₯)/γ€–sin⁑π‘₯ + cos〗⁑π‘₯ ) 𝑑π‘₯] + 𝐢1 = 1/2 [π‘₯βˆ’log⁑|sin⁑π‘₯+cos⁑π‘₯ |+𝐢2" " ] +𝐢1 = π‘₯/2βˆ’1/2 log⁑〖 |sin⁑π‘₯+cos⁑π‘₯ |γ€—+𝐢1+𝐢2/2 = 𝒙/𝟐 βˆ’πŸ/𝟐 π’π’π’ˆβ‘γ€– |π’”π’Šπ’β‘π’™+𝒄𝒐𝒔⁑𝒙 |γ€—+π‘ͺ

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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.