Slide3.JPG

Slide4.JPG
Slide5.JPG

  1. Chapter 7 Class 12 Integrals
  2. Concept wise

Transcript

Ex 7.10, 2 Evaluate the integrals using substitution ∫_0^(πœ‹/2)β–’γ€–βˆš(sin⁑〖" " Ο•" " γ€— ) cos^5⁑ϕ 𝑑ϕ〗 Let 𝐼=∫_0^(πœ‹/2)β–’γ€–βˆšπ‘ π‘–π‘›Ο• cos^5⁑〖ϕ 𝑑ϕ〗 γ€— 𝐼=∫_0^(πœ‹/2)β–’γ€–βˆšπ‘ π‘–π‘›Ο• cos^4⁑〖ϕ π‘π‘œπ‘ Ο• 𝑑ϕ〗 γ€— 𝐼=∫_0^(πœ‹/2)β–’γ€–βˆš(sin⁑ϕ ) (1βˆ’sin^2⁑ϕ )^2 π‘π‘œπ‘ Ο• 𝑑ϕ〗 Put 𝑑=sin⁑ϕ Differentiating w.r.t. Ο• 𝑑𝑑/𝑑ϕ=𝑑(sin⁑ϕ )/𝑑ϕ 𝑑𝑑/𝑑ϕ=cos⁑ϕ 𝑑𝑑/cos⁑ϕ =𝑑ϕ Hence, when Ο• varies form 0 to πœ‹/2 , 𝑑 varies form 0 to 1 Hence we can write the integral as 𝐼=∫_0^(πœ‹/2)β–’γ€–βˆšπ‘ π‘–π‘›Ο• (1βˆ’sin^2⁑ϕ )^2 π‘π‘œπ‘ Ο• 𝑑ϕ〗 =∫_0^1β–’γ€–βˆšπ‘‘ (1βˆ’π‘‘^2 )^2 π‘π‘œπ‘ Ο• 𝑑𝑑/π‘π‘œπ‘ Ο•γ€— =∫_0^1β–’γ€–βˆšπ‘‘ (1βˆ’π‘‘^2 )^2 𝑑𝑑〗 =∫_0^1▒〖𝑑^(1/2) (1βˆ’γ€–2𝑑〗^2+𝑑^4 )𝑑𝑑〗 =∫_0^1β–’γ€– (𝑑^(1/2)βˆ’γ€–2𝑑〗^(2 + 1/2)+𝑑^(4 + 1/2) ) 𝑑𝑑〗 =∫_0^1▒〖𝑑^(1/2) π‘‘π‘‘γ€—βˆ’2∫1▒𝑑^(3/2) 𝑑𝑑+∫1▒𝑑^(9/2) 𝑑𝑑 =[𝑑^(1/2 +1)/(1/2 +1)]_0^1βˆ’2[𝑑^(3/2 +1)/(3/2 +1)]_0^1+[𝑑^(9/2 +1)/(9/2 +1)]_0^1 =[𝑑^(3/2 )/(3/2)]_0^1βˆ’2[𝑑^(7/2 )/(7/2)]_0^1+[𝑑^(11/2 )/(11/2)]_0^1 =2/3 (1^(3/2)βˆ’0^(3/2) )βˆ’2 Γ— 2/7 (1^(7/2)βˆ’0^(7/2) )+2/11 [1^(11/2)βˆ’0^(11/2) ] =2/3βˆ’4/7+2/11 =(2 Γ— 7 Γ— 11 βˆ’ 4 Γ— 3 Γ— 11 + 2 Γ— 3 Γ—7)/(2 Γ— 7 Γ— 11) =(154 βˆ’ 132 + 42)/231 =πŸ”πŸ’/πŸπŸ‘πŸ

Chapter 7 Class 12 Integrals
Concept wise

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.