Example 7 - Chapter 7 Class 12 Integrals - Part 7

Example 7 - Chapter 7 Class 12 Integrals - Part 8
Example 7 - Chapter 7 Class 12 Integrals - Part 9

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.