Integration Full Chapter Explained -


Slide34.JPG Slide35.JPG Slide36.JPG

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


Misc 31 Evaluate the definite integral ∫_0^(πœ‹/2)β–’γ€–sin⁑2π‘₯ tan^(βˆ’1)⁑(sin⁑π‘₯ ) γ€— 𝑑π‘₯ ∫_0^(πœ‹/2)β–’γ€–sin⁑2π‘₯ tan^(βˆ’1)⁑(sin⁑π‘₯ ) γ€— 𝑑π‘₯ = ∫_0^(πœ‹/2)β–’γ€–2 sin⁑π‘₯ cos⁑π‘₯ tan^(βˆ’1)⁑(sin⁑π‘₯ ) γ€— 𝑑π‘₯ Let sin⁑π‘₯=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ cos⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/cos⁑π‘₯ Substituting x and dx ∫1_0^(πœ‹/2)β–’γ€–2 sin⁑〖π‘₯ cos⁑〖π‘₯ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (sin⁑〖π‘₯) γ€— γ€— γ€— γ€— 𝑑π‘₯ = ∫1_0^1β–’γ€–2𝑑 cos⁑〖π‘₯ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑) γ€— γ€— 𝑑𝑑/π‘π‘œπ‘ π‘₯ = ∫1_0^1β–’γ€–2𝑑 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑑 γ€— 𝑑𝑑 = 2∫1_0^1▒〖𝑑 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑) γ€— 𝑑𝑑 =2(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘‘βˆ«1▒𝑑 𝑑𝑑 βˆ’ ∫1β–’γ€–((𝑑 (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) ))/𝑑𝑑 ∫1▒〖𝑑 𝑑𝑑 γ€—) γ€— 𝑑𝑑) Now we know, ∫1▒〖𝑓 (π‘₯) 𝑔(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’γ€–((𝑑 𝑓 (π‘₯))/𝑑π‘₯ ∫1▒〖𝑔(π‘₯) 𝑑π‘₯γ€—) γ€— 𝑑π‘₯ Putting f(x) = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑) & g(x) = t = 2(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑑 (〖𝑑/2γ€—^2 )βˆ’βˆ«1β–’1/(1 + 𝑑^2 )×𝑑^2/2 𝑑𝑑) = 2(〖𝑑/2γ€—^2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘‘βˆ’1/2 ∫1▒𝑑^2/2 𝑑𝑑) = 𝑑^2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘‘βˆ’βˆ«1▒𝑑^2/(1 + 𝑑^2 ) 𝑑𝑑 Solving 𝑰_𝟏 I_1 = ∫1▒𝑑^2/(1 + 𝑑^2 ) 𝑑𝑑 Adding and Subtracting 1 in numerator. I_1 = ∫1β–’((𝑑^2 + 1 βˆ’ 1)/(𝑑^2 + 1))𝑑𝑑 I_1= ∫1β–’((𝑑^2 + 1)/(𝑑^2 + 1)βˆ’1/(𝑑^2 + 1))𝑑𝑑 I_1= ∫1β–’(1βˆ’1/(𝑑^2 + 1)) 𝑑𝑑 I_1= ∫1β–’γ€–π‘‘π‘‘βˆ’βˆ«1▒𝑑𝑑/(𝑑^2 + 1)γ€— I_1= t βˆ’ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (t) Thus, our equation becomes ∴ ∫1β–’γ€–γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑)×𝑑 𝑑𝑑〗= 𝑑^2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑑 βˆ’I_1 = 𝑑^2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘‘βˆ’(π‘‘βˆ’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑑) = 𝑑^2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘‘βˆ’π‘‘+γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑑 Now, 2∫1_0^1β–’γ€–γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑) 𝑑 𝑑𝑑〗 =[𝑑^2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘‘βˆ’π‘‘+γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 𝑑]_0^1 =(1^2Γ—γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 1βˆ’1+γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) 1)βˆ’(0βˆ’0+γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (0)) =(πœ‹/4βˆ’ 1+πœ‹/4)βˆ’0 = 𝝅/πŸβˆ’πŸ

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.