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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

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

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo