Misc 26 - Definite integral sinx cosx / cos4 x + sin4 x - Definate Integration - By Substitution

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  1. Chapter 7 Class 12 Integrals
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Misc 26 Evaluate the definite integral ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(cos^4⁑π‘₯ + sin^4⁑π‘₯ ) 𝑑π‘₯γ€— ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(cos^4⁑π‘₯ + sin^4⁑π‘₯ ) 𝑑π‘₯γ€— Adding and subtracting 2 〖𝑠𝑖𝑛〗^2 π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯ from denominator = ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(cos^4⁑π‘₯ + sin^4⁑〖π‘₯ + 2〖𝑠𝑖𝑛〗^2 π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯ βˆ’ 2〖𝑠𝑖𝑛〗^2 π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯γ€— ) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(γ€–γ€–(cosγ€—^2⁑π‘₯ + sin^2⁑π‘₯)γ€—^2 βˆ’2〖𝑠𝑖𝑛〗^2 π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(1 βˆ’ (4 〖𝑠𝑖𝑛〗^2 π‘₯ γ€–π‘π‘œπ‘ γ€—^2 π‘₯)/2) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(1 βˆ’ 1/2 ((2 sin⁑π‘₯ cos⁑π‘₯)/2)^2 ) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ cos⁑π‘₯)/(1 βˆ’ (〖𝑠𝑖𝑛〗^2 2π‘₯)/2) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–(2 sin⁑π‘₯ cos⁑π‘₯)/(2 βˆ’ 〖𝑠𝑖𝑛〗^(2 ) 2π‘₯) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–sin⁑2π‘₯/(1 + (1βˆ’ 〖𝑠𝑖𝑛〗^(2 ) 2π‘₯)) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–sin⁑2π‘₯/(1 + γ€–π‘π‘œπ‘ γ€—^(2 ) 2π‘₯) γ€— d⁑π‘₯ Let t = cos 2π‘₯ 𝑑𝑑/𝑑π‘₯=βˆ’2 sin⁑2π‘₯ (βˆ’π‘‘π‘‘)/2=sin⁑2π‘₯ 𝑑π‘₯ Substituting value and changing limits = (βˆ’1)/2 ∫1_1^0▒𝑑𝑑/(1 + 𝑑^2 ) = (βˆ’1)/2 [γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑)]_1^0 = (βˆ’1)/2 [γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (0)βˆ’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (1)] = (βˆ’1)/2 (0βˆ’πœ‹/4) = 𝝅/πŸ–

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