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Find Integration of sin x sin 2x sin 3x - Ex 7.3, 6 - NCERT Maths

Ex 7.3, 6 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.3, 6 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.3, 6 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.3, 6 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.3, 6 𝑠𝑖𝑛 π‘₯ sin⁑2π‘₯ sin⁑3π‘₯ ∫1β–’sin⁑〖π‘₯ sin⁑〖2π‘₯ sin⁑3π‘₯ γ€— γ€— 𝑑π‘₯ =∫1β–’γ€–(sin⁑π‘₯ sin⁑2π‘₯ ) sin⁑3π‘₯ γ€— 𝑑π‘₯ We know that 2 sin⁑𝐴 sin⁑𝐡=βˆ’cos⁑(𝐴+𝐡)+cos⁑(π΄βˆ’π΅) sin⁑𝐴 sin⁑𝐡=1/2 [βˆ’cos⁑(𝐴+𝐡)+cos⁑(π΄βˆ’π΅) ] sin⁑𝐴 sin⁑𝐡=1/2 [cos⁑(π΄βˆ’π΅)βˆ’cos⁑(𝐴+𝐡) ] Replace A by π‘₯ & B by 2π‘₯ sin⁑π‘₯ sin⁑2π‘₯=1/2 [cos⁑(π‘₯βˆ’2π‘₯)βˆ’cos⁑(π‘₯+2π‘₯) ] sin⁑π‘₯ sin⁑2π‘₯ =1/2 [cos⁑(βˆ’π‘₯)βˆ’cos⁑(3π‘₯) ] sin⁑π‘₯ sin⁑2π‘₯ =1/2 [cos⁑〖 π‘₯γ€—βˆ’cos⁑3π‘₯ ] Thus, our equation becomes ∫1▒𝐬𝐒𝐧⁑〖𝒙 π¬π’π§β‘πŸπ’™ sin⁑3π‘₯ γ€— 𝑑π‘₯ =∫1β–’γ€–πŸ/𝟐 (π’„π’π’”β‘π’™βˆ’π’„π’π’”β‘πŸ‘π’™ ) γ€— . sin⁑3π‘₯.𝑑π‘₯ =1/2 ∫1β–’(cos⁑π‘₯βˆ’cos⁑3π‘₯ ) sin⁑3π‘₯ 𝑑π‘₯ =1/2 [∫1β–’(cos⁑π‘₯. sin⁑3π‘₯βˆ’cos⁑3π‘₯. sin⁑3π‘₯ ) ]𝑑π‘₯ =1/2 [∫1β–’γ€–cos⁑π‘₯. sin⁑3π‘₯ γ€— 𝑑π‘₯βˆ’βˆ«1β–’γ€–cos⁑3π‘₯. sin⁑3π‘₯ γ€— 𝑑π‘₯] (βˆ΅π‘π‘œπ‘ β‘(βˆ’π‘₯)=π‘π‘œπ‘ β‘π‘₯) ∫1▒〖𝒄𝒐𝒔⁑𝒙. π’”π’Šπ’β‘πŸ‘π’™ γ€— 𝒅𝒙 We know that 2 𝑠𝑖𝑛⁑𝐴 π‘π‘œπ‘ β‘π΅ =𝑠𝑖𝑛⁑(𝐴+𝐡)+𝑠𝑖𝑛⁑(π΄βˆ’π΅) 𝑠𝑖𝑛⁑𝐴 π‘π‘œπ‘ β‘π΅=1/2 [𝑠𝑖𝑛⁑(𝐴+𝐡)+𝑠𝑖𝑛⁑(π΄βˆ’π΅) ] Replace A by 3π‘₯ & B by π‘₯ sin⁑3π‘₯ cos⁑π‘₯ = 1/2 [𝑠𝑖𝑛⁑(π‘₯+3π‘₯)+sin⁑(3π‘₯βˆ’π‘₯) ] = 1/2 [𝑠𝑖𝑛⁑4π‘₯+sin⁑2π‘₯ ] ∫1β–’γ€–π’„π’π’”β‘πŸ‘π’™. π’”π’Šπ’β‘πŸ‘π’™ γ€— 𝒅𝒙 We know that 2 𝑠𝑖𝑛⁑𝐴 π‘π‘œπ‘ β‘π΅ =𝑠𝑖𝑛⁑(𝐴+𝐡)+𝑠𝑖𝑛⁑(π΄βˆ’π΅) 𝑠𝑖𝑛⁑𝐴 π‘π‘œπ‘ β‘π΅ =1/2 [𝑠𝑖𝑛⁑(𝐴+𝐡)+𝑠𝑖𝑛⁑(π΄βˆ’π΅) ] Replace A by 3π‘₯ & B by 3π‘₯ sin⁑3π‘₯ cos⁑3π‘₯ = 1/2 [𝑠𝑖𝑛⁑(3π‘₯+3π‘₯)+sin⁑(3π‘₯βˆ’3π‘₯) ] = 1/2 [𝑠𝑖𝑛⁑6π‘₯+sin⁑0 ] =1/2 [𝑠𝑖𝑛⁑6π‘₯ ] Hence ∫1β–’γ€–sin⁑3π‘₯.cos⁑π‘₯ γ€— 𝑑π‘₯ =1/2 ∫1β–’[𝑠𝑖𝑛⁑4π‘₯+sin⁑2π‘₯ ] 𝑑π‘₯ Hence ∫1β–’γ€–cos⁑3π‘₯.sin⁑3π‘₯ γ€— 𝑑π‘₯ =1/2 ∫1β–’sin⁑6π‘₯ 𝑑π‘₯ Thus, our equation becomes ∫1β–’sin⁑〖π‘₯ sin⁑〖2π‘₯ sin⁑3π‘₯ γ€— γ€— 𝑑π‘₯ =1/2 [∫1β–’γ€–sin⁑3π‘₯ cos⁑3π‘₯ γ€— 𝑑π‘₯βˆ’βˆ«1β–’γ€–sin⁑3π‘₯ cos⁑3π‘₯ γ€— 𝑑π‘₯] =1/2 [1/2 ∫1β–’(sin⁑4π‘₯+sin⁑2π‘₯ ) 𝑑π‘₯βˆ’1/2 ∫1β–’(sin⁑6π‘₯ ) 𝑑π‘₯] =1/4 [∫1β–’(sin⁑4π‘₯+sin⁑2π‘₯ ) 𝑑π‘₯βˆ’βˆ«1β–’(sin⁑6π‘₯ ) 𝑑π‘₯] =1/4 [∫1β–’sin⁑4π‘₯ 𝑑π‘₯+∫1β–’sin⁑2π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’sin⁑6π‘₯ 𝑑π‘₯] ∫1β–’sin⁑(π‘Žπ‘₯+𝑏) 𝑑π‘₯=βˆ’π‘π‘œπ‘ β‘(π‘Žπ‘₯ + 𝑏)/π‘Ž +𝐢 =1/4 [(βˆ’cos⁑4π‘₯)/4 +(γ€–βˆ’cos〗⁑2π‘₯/2) βˆ’((βˆ’cos⁑6π‘₯)/6)]+𝐢 =1/4 [(βˆ’cos⁑4π‘₯)/4 βˆ’cos⁑2π‘₯/2+cos⁑6π‘₯/6]+𝐢 =𝟏/πŸ’ [π’„π’π’”β‘πŸ”π’™/πŸ” βˆ’π’„π’π’”β‘πŸ’π’™/πŸ’ βˆ’ π’„π’π’”β‘πŸπ’™/𝟐 ]+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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.