Integration Full Chapter Explained - Integration Class 12 - Everything you need


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


Example 7 Find (i) ∫1β–’cos^2⁑π‘₯ 𝑑π‘₯ ∫1β–’cos^2⁑π‘₯ 𝑑π‘₯ =∫1β–’((cos⁑2π‘₯ + 1)/2) 𝑑π‘₯ = 1/2 ∫1β–’(cos⁑2π‘₯+1) 𝑑π‘₯ = 1/2 [∫1β–’cos⁑2π‘₯ 𝑑π‘₯+∫1β–’1 𝑑π‘₯] We know cos⁑2π‘₯=2 cos^2⁑π‘₯βˆ’1 cos⁑2π‘₯+1=2 cos^2⁑π‘₯ (cos⁑2π‘₯ + 1)/2=cos^2⁑π‘₯ ∫1β–’π’„π’π’”β‘πŸπ’™ 𝒅𝒙 Let 2π‘₯=𝑑 2 =𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=1/2 𝑑𝑑 ∫1β–’cos⁑𝑑 . 1/2 𝑑𝑑 =1/2 (sin⁑𝑑+𝐢1) Putting value of 𝑑 = 2π‘₯ =1/2 sin⁑2π‘₯+𝐢1 ∫1β–’πŸ 𝒅𝒙 =∫1β–’π‘₯^0 𝑑π‘₯ =[π‘₯^(0 + 1)/(0 + 1)]+𝐢 =π‘₯+𝐢2 Thus, ∫1β–’cos^2⁑π‘₯ 𝑑π‘₯=1/2 [∫1β–’cos⁑2π‘₯ 𝑑π‘₯+∫1β–’1 𝑑π‘₯] =1/2 [1/2 sin⁑2π‘₯+𝐢1+π‘₯+𝐢2] =1/4 sin⁑2π‘₯+π‘₯/2+1/2(𝐢1+𝐢2) =𝒙/𝟐+𝟏/πŸ’ π¬π’π§β‘πŸπ’™+π‘ͺ ("From (1) and (2) " ) ("Let" 𝐢1+𝐢2=𝐢) Example 7 Find (ii) ∫1β–’γ€–sin⁑〖2π‘₯ γ€— cos⁑3π‘₯ γ€— 𝑑π‘₯ We know that 2 sin⁑𝐴 cos⁑𝐡=sin⁑(𝐴+𝐡)+sin⁑(π΄βˆ’π΅) sin⁑𝐴 cos⁑𝐡=1/2 [sin⁑(𝐴+𝐡)+sin⁑(π΄βˆ’π΅) ] Replace A by 2π‘₯ & B by 3π‘₯ sin⁑2π‘₯ cos⁑3π‘₯=1/2 [sin⁑(2π‘₯+3π‘₯)+sin⁑(2π‘₯βˆ’3π‘₯) ] sin⁑2π‘₯ cos⁑3π‘₯=1/2 [sin⁑(5π‘₯)+sin⁑(βˆ’π‘₯) ] sin⁑2π‘₯ cos⁑3π‘₯=1/2 [sin⁑5π‘₯βˆ’sin⁑π‘₯ ] ∫1β–’(sin⁑〖2π‘₯ γ€— cos⁑3π‘₯ ) 𝑑π‘₯=1/2 ∫1β–’(sin⁑5π‘₯βˆ’sin⁑π‘₯ ) 𝑑π‘₯ =1/2 [∫1β–’sin⁑5π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’sin⁑π‘₯ 𝑑π‘₯] ∫1β–’π’”π’Šπ’β‘πŸ“π’™ 𝒅𝒙 Let 5π‘₯=𝑑 5 = 𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=1/5 𝑑𝑑 =∫1β–’sin⁑𝑑 . 1/5 𝑑𝑑 =1/5 ∫1β–’sin⁑𝑑 . 𝑑𝑑 =1/5 (γ€–βˆ’cos〗⁑𝑑+𝐢1) =βˆ’1/5 cos⁑𝑑+1/5 𝐢1 Putting value of 𝑑 =βˆ’1/5 cos⁑5π‘₯+1/5 𝐢1 ∫1β–’sin⁑π‘₯ 𝑑π‘₯ =βˆ’cos⁑π‘₯+𝐢2 Thus, ∫1β–’γ€–sin⁑〖2π‘₯ γ€— cos⁑3π‘₯ γ€—=1/2 [∫1β–’γ€–sin⁑5π‘₯ 𝑑π‘₯γ€—βˆ’βˆ«1β–’sin⁑π‘₯ 𝑑π‘₯] =1/2 [βˆ’1/5 cos⁑5π‘₯+1/5 𝐢1βˆ’(βˆ’cos⁑π‘₯ )+𝐢2] =1/2 [βˆ’1/5 cos⁑5π‘₯+cos⁑π‘₯+1/5 𝐢1βˆ’πΆ2] =1/2 [βˆ’1/5 cos⁑5π‘₯+cos⁑π‘₯+𝐢] =(βˆ’πŸ)/𝟏𝟎 πœπ¨π¬β‘πŸ“π’™+𝟏/𝟐 πœπ¨π¬β‘π’™+π‘ͺ 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
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.