Integration Full Chapter Explained -



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


Ex 7.3, 10 Integrate the function 𝑠𝑖𝑛4 π‘₯ ∫1β–’sin^4⁑π‘₯ 𝑑π‘₯ =∫1β–’(sin^2⁑π‘₯ )^2 𝑑π‘₯ =∫1β–’((1 βˆ’ cos⁑2π‘₯)/2)^2 𝑑π‘₯ =1/4 ∫1β–’(1βˆ’cos⁑2π‘₯ )^2 𝑑π‘₯ We know that π‘π‘œπ‘ β‘2πœƒ=1βˆ’2 〖𝑠𝑖𝑛〗^2β‘πœƒ 2 〖𝑠𝑖𝑛〗^2β‘πœƒ=1βˆ’π‘π‘œπ‘ β‘2πœƒ 〖𝑠𝑖𝑛〗^2β‘πœƒ=(1 βˆ’ π‘π‘œπ‘ β‘2πœƒ)/2 Replace πœƒ by π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯=(1 βˆ’ π‘π‘œπ‘ β‘2π‘₯)/2 =1/4 ∫1β–’(1^2+(cos⁑2π‘₯ )^2βˆ’2(1)(cos⁑2π‘₯ )) 𝑑π‘₯ =1/4 ∫1β–’(1+cos^2⁑2π‘₯βˆ’2 cos⁑2π‘₯ ) 𝑑π‘₯ =1/4 ∫1β–’(1+(1 + cos⁑4π‘₯)/2βˆ’2 cos⁑2π‘₯ ) 𝑑π‘₯ =1/4 ∫1β–’1 𝑑π‘₯+1/8 ∫1β–’(1+cos⁑4π‘₯ ) 𝑑π‘₯βˆ’2/4 ∫1β–’cos⁑2π‘₯ 𝑑π‘₯ =1/4 ∫1β–’1 𝑑π‘₯+1/8 ∫1β–’1 𝑑π‘₯+1/8 ∫1β–’cos⁑4π‘₯ 𝑑π‘₯βˆ’1/2 ∫1β–’cos⁑2π‘₯ 𝑑π‘₯ We know that cos⁑2πœƒ=2 cos^2β‘γ€–πœƒβˆ’1γ€— cos⁑〖2πœƒ+1γ€—=2 cos^2β‘πœƒ Replace πœƒ by 2π‘₯ cos⁑〖4π‘₯+1γ€—=2 cos^2⁑2π‘₯ cos⁑〖4π‘₯ + 1γ€—/2=cos^2⁑2π‘₯ ∫1β–’cos⁑(π‘Žπ‘₯+𝑏) 𝑑π‘₯=𝑠𝑖𝑛⁑(π‘Žπ‘₯ + 𝑏)/π‘Ž =π‘₯/4 + π‘₯/8 + 1/8 sin⁑4π‘₯/4 βˆ’ 1/2 sin⁑2π‘₯/2 +𝐢 =(2π‘₯ + π‘₯)/8 + 1/32 sin⁑4π‘₯βˆ’ 1/4 sin⁑2π‘₯+𝐢 =πŸ‘π’™/πŸ–βˆ’ 𝟏/πŸ’ π’”π’Šπ’β‘πŸπ’™++ 𝟏/πŸ‘πŸ π’”π’Šπ’β‘πŸ’π’™+π‘ͺ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.