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Misc 30 - Definite integral 0 -> pi/4 sin x + cos x - Miscellaneous

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Misc 30 Evaluate the definite integral ∫_0^(πœ‹/4)β–’γ€–(sin⁑π‘₯ + cos⁑π‘₯)/(9 + 16 sin⁑2π‘₯ ) γ€— 𝑑π‘₯ ∫_0^(πœ‹/4)β–’γ€– (sin⁑π‘₯ + cos⁑π‘₯)/(9 +16 sin⁑2π‘₯ ) γ€— 𝑑π‘₯ Let sin⁑π‘₯βˆ’cos⁑π‘₯=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ cos⁑π‘₯+sin⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(sin⁑π‘₯ + cos⁑π‘₯ ) Now, sin⁑π‘₯βˆ’cos⁑π‘₯=𝑑 Squaring both sides (sin⁑π‘₯βˆ’cos⁑π‘₯ )^2=𝑑^2 sin^2⁑π‘₯+cos^2⁑π‘₯βˆ’2 sin⁑π‘₯ cos⁑π‘₯=𝑑^2 1βˆ’2 sin⁑π‘₯ cos⁑π‘₯=𝑑^2 1βˆ’sin⁑2π‘₯=𝑑^2 1βˆ’π‘‘^2=sin⁑2π‘₯ sin⁑0βˆ’cos⁑0 =0βˆ’1=βˆ’1 Putting the values of dx and sin⁑2x, we get ∫_0^(πœ‹/4)β–’γ€– (sin⁑π‘₯ + cos⁑π‘₯)/(9 +16 sin⁑2π‘₯ ) γ€— 𝑑π‘₯=∫1_(βˆ’1)^0β–’γ€–(sin⁑π‘₯ + cos⁑π‘₯)/(9 +16 sin⁑2π‘₯ )×𝑑𝑑/(sin⁑π‘₯ + cos⁑π‘₯ )γ€— =∫_(βˆ’1)^0β–’γ€– 1/(9 +16 (1 βˆ’ 𝑑^2 ) ) γ€—. 𝑑𝑑 =∫_(βˆ’1)^0β–’γ€– 1/(9 +16 βˆ’ 16𝑑^2 ) γ€—. 𝑑𝑑 =∫_(βˆ’1)^0β–’γ€– 1/(25 βˆ’ 16𝑑^2 ) γ€—. 𝑑𝑑 =1/16 ∫_(βˆ’1)^0β–’γ€– 1/(25/16 βˆ’ 𝑑^2 ) γ€—. 𝑑𝑑 =1/16 ∫_(βˆ’1)^0β–’γ€– 1/((5/4)^2 βˆ’ 𝑑^2 ) γ€—. 𝑑𝑑 It is of form βˆ«β–’γ€–π‘‘π‘₯/(π‘Ž^2 βˆ’ π‘₯^2 ) = 1/2π‘Ž γ€–log 〗⁑|(π‘Ž + π‘₯)/(π‘Ž βˆ’ π‘₯)|+𝐢〗 Replacing π‘₯ by t and a by 5/4 , we get = γ€–1/16 [1/(2 . 5/4) γ€– log〗⁑〖 |( 5/4 + 𝑑)/( 5/4 βˆ’ 𝑑)|γ€— ]γ€—_(βˆ’1)^0 = γ€–1/4 [1/10 γ€– log〗⁑〖 |( 5 + 4𝑑)/( 5 βˆ’ 4𝑑)|γ€— ]γ€—_(βˆ’1)^0 = 1/40 γ€–log⁑〖 |( 5 + 4𝑑)/( 5 βˆ’ 4𝑑)|γ€—γ€—_(βˆ’1)^0 = 1/40 [log⁑|( 5 + 4(0))/( 5 βˆ’ 4(0) )|βˆ’log⁑|( 5 + 4(βˆ’1))/( 5 βˆ’ 4(βˆ’1) )| ] = 1/40 [log⁑|( 5 + 0)/( 5 βˆ’ 0)|βˆ’log⁑|( 5 βˆ’ 4)/( 5 + 4)| ] = 1/40 [log⁑〖5/5γ€—βˆ’log⁑〖1/9γ€— ] = 1/40 [γ€–log 〗⁑1βˆ’log⁑(1/9) ] = 1/40 (log⁑1+log⁑〖(1/9)^(βˆ’1) γ€— ) = 1/40 (0+log⁑〖9)γ€— = 𝟏/πŸ’πŸŽ π’π’π’ˆ πŸ— ("n log m = log " π‘š^𝑛 " " ) (log⁑〖1=0γ€— )

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.