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Ex 7.3, 5 - Integrate  sin3 x cos3 x - Chapter 7 Class 12

Ex 7.3, 5 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.3, 5 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.3, 5 Integrate sin^3⁑π‘₯ cos^3 π‘₯ ∫1β–’γ€–sin^3⁑π‘₯ cos^3 π‘₯γ€— 𝑑π‘₯ =∫1▒〖𝑠𝑖𝑛⁑π‘₯. sin^2⁑π‘₯ cos^3 π‘₯γ€— 𝑑π‘₯ =∫1▒〖𝑠𝑖𝑛⁑π‘₯ (1βˆ’cos^2⁑π‘₯ ) cos^3 π‘₯γ€— 𝑑π‘₯ =∫1β–’γ€–(1βˆ’cos^2⁑π‘₯ ) cos^3 π‘₯γ€—. 𝑠𝑖𝑛⁑π‘₯ 𝑑π‘₯ Let cos⁑π‘₯=𝑑 Differentiating w.r.t.x βˆ’sin⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’sin⁑π‘₯ ) (〖𝑠𝑖𝑛〗^2β‘πœƒ=1βˆ’γ€–π‘π‘œπ‘ γ€—^2β‘πœƒ) …(1) Thus, our equation becomes ∫1β–’γ€–sin^2⁑π‘₯ cos^3 π‘₯γ€— 𝑑π‘₯ =∫1β–’γ€–(1βˆ’cos^2⁑π‘₯ ) cos^3 π‘₯γ€—. 𝑠𝑖𝑛⁑π‘₯ 𝑑π‘₯" " =∫1β–’γ€–(1βˆ’π‘‘^2 ) 𝑑^3 γ€—. 𝑠𝑖𝑛⁑π‘₯Γ—1/(βˆ’sin⁑π‘₯ ) 𝑑𝑑" " =∫1β–’γ€–βˆ’(1βˆ’π‘‘^2 ) 𝑑^3 γ€— 𝑑𝑑" " =βˆ’βˆ«1β–’γ€–(𝑑^3βˆ’π‘‘^5 ) γ€— 𝑑𝑑" " =βˆ’[∫1▒𝑑^3 π‘‘π‘‘βˆ’βˆ«1▒𝑑^5 𝑑𝑑] =βˆ’[𝑑^(3 + 1)/(3 + 1) βˆ’ 𝑑^(5 + 1)/(5 + 1)]+𝐢 =βˆ’[𝑑^4/4 βˆ’ 𝑑^6/6]+𝐢 =γ€–βˆ’π‘‘γ€—^4/4 + 𝑑^6/6+𝐢 =𝑑^6/6 βˆ’ 𝑑^4/4 +𝐢 Putting back value of 𝑑=π‘π‘œπ‘ β‘π‘₯ =(〖𝒄𝒐𝒔〗^πŸ” 𝒙)/πŸ” βˆ’ (〖𝒄𝒐𝒔〗^πŸ’ 𝒙)/πŸ’ +π‘ͺ

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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.