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

Transcript

Ex 7.9, 17 ∫_0^(πœ‹/4)β–’(2 sec^2⁑π‘₯+π‘₯^3+2) 𝑑π‘₯ Let F(π‘₯)=∫1β–’γ€–(2 sec^2⁑〖π‘₯+π‘₯^3+2γ€— ) 𝑑π‘₯γ€— =2∫1β–’γ€–sec^2⁑〖π‘₯ 𝑑π‘₯γ€—+∫1β–’γ€–π‘₯^3 𝑑π‘₯+∫1β–’γ€–2 𝑑π‘₯γ€—γ€—γ€— =2 tan⁑〖π‘₯+π‘₯^4/4+2π‘₯γ€— Hence F(π‘₯)=2 tan⁑〖π‘₯+π‘₯^4/4+2π‘₯γ€— Now, ∫_0^(πœ‹/4)β–’γ€–(2𝑠𝑒𝑐^2 π‘₯+π‘₯^3+2)𝑑π‘₯=𝐹(πœ‹/4)βˆ’πΉ(0) γ€— =[2π‘‘π‘Žπ‘› πœ‹/4+(πœ‹/4)^4/4+2 πœ‹/4]βˆ’[2π‘‘π‘Žπ‘›(0)+(0)^4/4+2 Γ—0] =2π‘‘π‘Žπ‘› πœ‹/4+πœ‹^4/4^4 Γ— 1/4+πœ‹/2βˆ’[2Γ—0+0+0] =2 Γ—1+πœ‹^4/4^5 +πœ‹/2βˆ’0 =𝟐+𝝅^πŸ’/πŸπŸŽπŸπŸ’+𝝅/𝟐

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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.