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Misc 31 - Definite integral sin 2x tan-1 (sin x) dx - Miscellaneous

Misc 31 - Chapter 7 Class 12 Integrals - Part 2
Misc 31 - Chapter 7 Class 12 Integrals - Part 3
Misc 31 - Chapter 7 Class 12 Integrals - Part 4
Misc 31 - Chapter 7 Class 12 Integrals - Part 5

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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

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.