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Ex 7.2, 37 - Integrate x^3 sin (tan-1 x^4) / 1 + x^8 - Teachoo

Ex 7.2, 37 - Chapter 7 Class 12 Integrals - Part 2


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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.