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Ex 7.11, 20 - Value of (x3 + x cos x + tan5 x + 1) dx - Ex 7.11

Ex 7.11, 20 - Chapter 7 Class 12 Integrals - Part 2

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Ex 7.11, 20 The value of ∫_((βˆ’ πœ‹)/2)^(πœ‹/2)β–’γ€–(π‘₯^3+π‘₯ π‘π‘œπ‘  π‘₯+tan^5⁑〖π‘₯+1γ€— ) 𝑑π‘₯γ€— (A) 0 (B) 2 (C) πœ‹ (D) 1 ∫_((βˆ’πœ‹)/2)^(πœ‹/2)β–’γ€– (π‘₯^3+π‘₯ π‘π‘œπ‘  π‘₯+tan^5⁑〖π‘₯+1γ€— ) 𝑑π‘₯γ€— We know that ∫_(βˆ’π‘Ž)^(+π‘Ž)▒〖𝑓(π‘₯) 𝑑π‘₯γ€—={β–ˆ(0, 𝑖𝑓 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯)@&2∫_0^π‘Žβ–’γ€–π‘“(π‘₯) 𝑑π‘₯γ€—, 𝑖𝑓 𝑓(βˆ’π‘₯)=𝑓(π‘₯) )─ Thus, our equation becomes ∫_((βˆ’ πœ‹)/2)^(πœ‹/2)β–’γ€–(π‘₯^3+π‘₯ π‘π‘œπ‘  π‘₯+tan^5⁑〖π‘₯+1γ€— ) 𝑑π‘₯γ€— = ∫_0^(πœ‹/2)β–’γ€–0 𝑑π‘₯+∫_0^(πœ‹/2)β–’γ€–0 𝑑π‘₯+∫_0^(πœ‹/2)β–’γ€–0 𝑑π‘₯+𝟐∫_𝟎^(𝝅/𝟐)▒𝒅𝒙〗〗〗 = 2∫_0^(πœ‹/2)▒𝑑π‘₯ = 2 [π‘₯]_0^(πœ‹/2) = 2(πœ‹/2βˆ’0) =𝝅 Thus, correct option is C

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