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

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

Transcript

Ex 7.2, 37 Integrate (π‘₯3 𝑠𝑖𝑛 (tan^(βˆ’1)⁑〖π‘₯^4 γ€—))/(1 + π‘₯8) ∫1β–’γ€–" " (π‘₯3 𝑠𝑖𝑛 (tan^(βˆ’1)⁑〖π‘₯^4 γ€— ))/(1 + π‘₯8)γ€— . 𝑑π‘₯ Let tan^(βˆ’1)⁑〖π‘₯^4 γ€—= 𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 1/(1 +(π‘₯^4 )^2 ). 𝑑(π‘₯^4 )/𝑑π‘₯= 𝑑𝑑/𝑑π‘₯ 1/(1 +π‘₯^8 ). 4π‘₯^3=𝑑𝑑/𝑑π‘₯ (4π‘₯^3)/(1 + π‘₯^8 )=𝑑𝑑/𝑑π‘₯ (Using (𝑑(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑π‘₯))/𝑑π‘₯=1/(1 + π‘₯^2 ) and chain rule ) 𝑑π‘₯=(1 + π‘₯^8)/(4π‘₯^3 ) . 𝑑𝑑 Now, our function becomes ∫1β–’γ€–" " (π‘₯3 𝑠𝑖𝑛 (tan^(βˆ’1)⁑〖π‘₯^4 γ€— ))/(1 + π‘₯8)γ€— . 𝑑π‘₯ Putting γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑〖π‘₯^4 γ€—=𝑑 & 𝑑π‘₯=(1 + π‘₯^8)/(4π‘₯^3 ) . 𝑑𝑑 = ∫1β–’γ€–" " (π‘₯^3 𝑠𝑖𝑛 (𝑑))/(1 + π‘₯^8 )γ€—. (1 + π‘₯^8)/(4π‘₯^3 ) 𝑑𝑑" " = ∫1β–’γ€–" " sin⁑𝑑/4γ€— 𝑑𝑑" " = 1/4 ∫1β–’sin⁑𝑑 . 𝑑𝑑" " = (βˆ’1)/4 cos⁑𝑑+ 𝐢 = (βˆ’πŸ)/πŸ’ 〖𝒄𝒐𝒔 〗⁑(γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑〖𝒙^πŸ’ γ€— )+π‘ͺ (Using 𝑑=γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑〖π‘₯^4 γ€—)

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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.