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

Transcript

Example 21 Find ∫1▒𝑒^π‘₯ sin⁑π‘₯ 𝑑π‘₯ Let I1 = ∫1β–’γ€– 𝑒^π‘₯ γ€— sin⁑π‘₯ 𝑑π‘₯ I1 = sin⁑π‘₯ ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€—βˆ’βˆ«1β–’(𝑑(sin⁑π‘₯ )/𝑑π‘₯ ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€—) 𝑑π‘₯ I1 = 𝑒^π‘₯ sin⁑π‘₯βˆ’βˆ«1β–’γ€–cos⁑π‘₯ . 𝑒^π‘₯ 𝑑π‘₯γ€— Now we know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓′(π‘₯)∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = sin x and g(x) = ex Solving I2 I2 = ∫1β–’γ€–cos⁑π‘₯ . 𝑒^π‘₯ 𝑑π‘₯γ€— I2 = cos x ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€— – ∫1β–’γ€–((cos⁑π‘₯)β€²γ€— ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€—)𝑑π‘₯ I2 = cos x 𝑒^π‘₯ – ∫1β–’γ€–(βˆ’sin⁑π‘₯)γ€— 𝑒^π‘₯ 𝑑π‘₯ I2 = 𝑒^π‘₯ cos x + ∫1β–’sin⁑π‘₯ 𝑒^π‘₯ 𝑑π‘₯ I2 = 𝑒^π‘₯ cos x + 𝐼1 Now we know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓′(π‘₯)∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = sin x and g(x) = ex Now, Putting value of I2 in (1) , I1 = " " 𝑒^π‘₯ sin⁑π‘₯βˆ’βˆ«1β–’γ€–cos⁑π‘₯ 𝑒^π‘₯ γ€— 𝑑π‘₯ I1 = " " 𝑒^π‘₯ sin⁑π‘₯βˆ’(𝑒^π‘₯ cos⁑π‘₯+𝐼1)+𝐢 I1 = " " 𝑒^π‘₯ sin⁑π‘₯βˆ’π‘’^π‘₯ cos⁑π‘₯βˆ’πΌ1+𝐢 2I1 = " " 𝑒^π‘₯ sin⁑π‘₯βˆ’π‘’^π‘₯ cos⁑π‘₯ + 𝐢 I1 = 1/2 (𝑒^π‘₯ sin⁑π‘₯βˆ’π‘’^π‘₯ cos⁑π‘₯ ) + C π‘°πŸ = 𝒆^𝒙/𝟐 (π’”π’Šπ’β‘π’™βˆ’π’„π’π’”β‘π’™ ) + C

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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.