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Ex 7.11, 11 - Evaluate definite integral sin2 x dx - Ex 7.11

Ex 7.11, 11 - Chapter 7 Class 12 Integrals - Part 2


Transcript

Ex 7.11, 11 By using the properties of definite integrals, evaluate the integrals : ∫_((βˆ’ πœ‹)/2)^(πœ‹/2)β–’γ€– sin^2〗⁑π‘₯ 𝑑π‘₯ This is of form ∫_(βˆ’π‘Ž)^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯ where 𝑓(π‘₯)=sin^2⁑π‘₯ 𝑓(βˆ’π‘₯)=sin^2⁑(βˆ’π‘₯)=(βˆ’π‘ π‘–π‘›π‘₯)^2=sin^2⁑π‘₯ ∴ 𝑓(π‘₯)=𝑓(βˆ’π‘₯) Using the Property ∫_(βˆ’π‘Ž)^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=2,γ€— ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯ γ€— if f(βˆ’π‘₯)=𝑓(π‘₯) ∴ ∫_((βˆ’πœ‹)/2)^(πœ‹/2)β–’γ€–sin^2⁑〖π‘₯ 𝑑π‘₯γ€—=2∫_0^(πœ‹/2)β–’γ€–γ€–π’”π’Šπ’γ€—^𝟐 𝒙 𝑑π‘₯γ€—γ€— =2∫_0^(πœ‹/2)β–’[(𝟏 βˆ’ π’„π’π’”β‘πŸπ’™)/𝟐]𝑑π‘₯ =∫_0^(πœ‹/2)β–’γ€–(1βˆ’cos⁑2π‘₯ ) 𝑑π‘₯γ€— = [π‘₯ βˆ’sin⁑2π‘₯/2]_0^(πœ‹/2) = [πœ‹/2βˆ’sin⁑2(πœ‹/2)/2]βˆ’ [0βˆ’sin⁑〖2(0)γ€—/2] = πœ‹/2βˆ’sinβ‘πœ‹/2βˆ’0 = πœ‹/2βˆ’0+0 = 𝝅/𝟐 ∡ cos 2x = 1 βˆ’ 2 〖𝑠𝑖𝑛〗^2 π‘₯ β‡’ 2 〖𝑠𝑖𝑛〗^2 π‘₯ = 1 βˆ’ cos 2x β‡’ 〖𝑠𝑖𝑛〗^2 π‘₯ = "1 βˆ’ cos 2x" /2

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Davneet Singh

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.