Integration Full Chapter Explained - Integration Class 12 - Everything you need

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.