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Misc 26 - Definite integral sinx cosx / cos4 x + sin4 x - Miscellaneou

Misc 26 - Chapter 7 Class 12 Integrals - Part 2
Misc 26 - Chapter 7 Class 12 Integrals - Part 3 Misc 26 - Chapter 7 Class 12 Integrals - Part 4


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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 ) 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 + (1 βˆ’ 〖𝑠𝑖𝑛〗^(2 ) 2π‘₯)) 𝑑π‘₯γ€— = ∫_0^(πœ‹/4)β–’γ€–sin⁑2π‘₯/(1 + γ€–π‘π‘œπ‘ γ€—^2 2π‘₯) γ€— 𝑑π‘₯ (γ€–π‘π‘œπ‘ γ€—^2 π‘₯+〖𝑠𝑖𝑛〗^2 π‘₯=1" " ) ("2 sin " π‘₯ cos⁑π‘₯=sin⁑2π‘₯ " " ) ("2 sin " π‘₯ cos⁑π‘₯=sin⁑2π‘₯ " " ) Let t = cos 2π‘₯ 𝑑𝑑/𝑑π‘₯=βˆ’2 sin⁑2π‘₯ (βˆ’π‘‘π‘‘)/2=sin⁑2π‘₯ 𝑑π‘₯ So, our equation becomes ∫_0^(πœ‹/4)β–’γ€–sin⁑2π‘₯/(1 + γ€–π‘π‘œπ‘ γ€—^2 2π‘₯) γ€— 𝑑π‘₯ = (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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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.