Example 7 - Chapter 7 Class 12 Integrals - Part 7

Advertisement

Example 7 - Chapter 7 Class 12 Integrals - Part 8

Advertisement

Example 7 - Chapter 7 Class 12 Integrals - Part 9

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Example 7 Find (iii) ∫1β–’sin^3⁑π‘₯ 𝑑π‘₯ We know that 𝑠𝑖𝑛 3π‘₯=3 𝑠𝑖𝑛⁑π‘₯βˆ’4 〖𝑠𝑖𝑛〗^3⁑π‘₯ 4 〖𝑠𝑖𝑛〗^3 π‘₯=3 𝑠𝑖𝑛⁑π‘₯βˆ’π‘ π‘–π‘›β‘3π‘₯ 〖𝑠𝑖𝑛〗^3 π‘₯=(3 𝑠𝑖𝑛⁑π‘₯ βˆ’ 𝑠𝑖𝑛⁑3π‘₯)/4 ∫1β–’sin^3⁑π‘₯ 𝑑π‘₯=∫1β–’(3 sin⁑π‘₯ βˆ’ sin⁑3π‘₯)/4 𝑑π‘₯ =1/4 ∫1β–’(3 sin⁑π‘₯βˆ’sin⁑3π‘₯ ) 𝑑π‘₯ =1/4 [3∫1β–’sin⁑π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’sin⁑3π‘₯ 𝑑π‘₯] ∫1β–’π’”π’Šπ’β‘πŸ‘π’™ 𝒅𝒙 Let 3π‘₯=𝑑 3=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=1/3 𝑑𝑑 ∫1β–’π’”π’Šπ’β‘πŸ‘π’™ 𝒅𝒙=∫1β–’sin⁑𝑑 . 1/3 𝑑𝑑 =1/3 ∫1β–’sin⁑𝑑 . 𝑑𝑑 =1/3 (γ€–βˆ’cos〗⁑𝑑+𝐢1) =βˆ’1/3 cos⁑𝑑+1/3 𝐢1 Putting value of 𝑑 =βˆ’1/3 cos⁑3π‘₯+1/3 𝐢1 ∫1β–’π’”π’Šπ’β‘π’™ 𝒅𝒙 =βˆ’cos⁑π‘₯+𝐢2 Thus, ∫1β–’sin^3⁑π‘₯ 𝑑π‘₯=1/4 [3∫1β–’γ€–sin⁑π‘₯ 𝑑π‘₯γ€—βˆ’βˆ«1β–’sin⁑3π‘₯ 𝑑π‘₯] =1/4 [3(βˆ’cos⁑π‘₯+𝐢2)βˆ’(βˆ’1/3 cos⁑3π‘₯+1/3 𝐢1)] =1/4 [βˆ’3 cos⁑π‘₯+3 𝐢2+ 1/3 cos⁑3π‘₯+1/3 𝐢1] =1/4 [βˆ’3 cos⁑π‘₯+1/3 cos⁑3π‘₯+(3 𝐢2βˆ’1/3 𝐢1)] =(βˆ’3)/4 cos⁑π‘₯+1/12 cos⁑3π‘₯+1/4 (3 𝐢2βˆ’1/3 𝐢1) =(βˆ’πŸ‘)/πŸ’ 𝒄𝒐𝒔⁑𝒙+𝟏/𝟏𝟐 π’„π’π’”β‘πŸ‘π’™+π‘ͺ ("As" 1/4 (3 𝐢2βˆ’1/3 𝐢1)=𝐢)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.