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Ex 7.10, 2 - Evaluate using substitution: root sin cos 5

Ex 7.10, 2 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.10, 2 - Chapter 7 Class 12 Integrals - Part 3


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

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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.