Misc 30 - Definite integral 0 -> pi/4 sin x + cos x - Definate Integration - By Substitution

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  1. Chapter 7 Class 12 Integrals
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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‘₯ ) Again 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‘₯ Putting the values of dx and sin⁑2x, we get βˆ«β–’γ€–(sin⁑π‘₯ + cos⁑π‘₯)/(9 +16 sin⁑2π‘₯ ) γ€— 𝑑π‘₯=βˆ«β–’γ€–(sin⁑π‘₯ + cos⁑π‘₯)/(9 +16 sin⁑2π‘₯ ) 〗×𝑑𝑑/(sin⁑π‘₯ + cos⁑π‘₯ ) =∫_(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 = 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 〗⁑(1)βˆ’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 β‘γ€–πŸ— )γ€—

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