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Ncert Solution Class 12 Maths Chapter 7 Miscellaneous Exercise

Misc 28 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 23, 2019 by


Transcript

Misc 28 Evaluate the definite integral โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€–(sinโก๐‘ฅ + cosโก๐‘ฅ)/โˆš(sinโกใ€–2๐‘ฅ ใ€— ) ใ€— โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€–(sinโก๐‘ฅ + cosโก๐‘ฅ)/โˆš(sinโกใ€–2๐‘ฅ ใ€— ) ๐‘‘๐‘ฅ ใ€— = โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (sinโก๐‘ฅ + cosโก๐‘ฅ)/โˆš(2 sinโก๐‘ฅ cosโก๐‘ฅ ) ๐‘‘๐‘ฅ ใ€— = โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (sinโก๐‘ฅ/โˆš(2 sinโก๐‘ฅ cosโก๐‘ฅ )+cosโก๐‘ฅ/โˆš(2 sinโก๐‘ฅ cosโก๐‘ฅ )) ๐‘‘๐‘ฅ ใ€— = โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (โˆš(sinโก๐‘ฅ )/(โˆš2 โˆš(cosโก๐‘ฅ ))+โˆš(cosโก๐‘ฅ )/(โˆš2 โˆš(sinโก๐‘ฅ ))) ๐‘‘๐‘ฅ ใ€— ("Using" sinโก2๐œƒ=2 sinโก๐œƒ cosโก๐œƒ ) = โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (1/โˆš2 โˆš(sinโก๐‘ฅ/cosโก๐‘ฅ ) + 1/โˆš2 โˆš(cosโก๐‘ฅ/sinโก๐‘ฅ )) ๐‘‘๐‘ฅ ใ€— = 1/โˆš2 โˆซ_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€–(โˆš(tanโก๐‘ฅ )+โˆš(cotโก๐‘ฅ ) ) ๐‘‘๐‘ฅ ใ€— = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– [โˆš(cotโก๐‘ฅ )+1/โˆš(cotโก๐‘ฅ )] ใ€— ๐‘‘๐‘ฅ = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’(cotโก๐‘ฅ + 1)/โˆš(cotโก๐‘ฅ ) ๐‘‘๐‘ฅ = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€–โˆš(tanโก๐‘ฅ ) (cotโก๐‘ฅ+1) ใ€— ๐‘‘๐‘ฅ ("Using" ๐‘ก๐‘Ž๐‘›โก๐‘ฅ= 1/๐‘๐‘œ๐‘กโก๐‘ฅ ) Let tanโก๐‘ฅ=๐‘ก^2 Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. sec^2 ๐‘ฅ=2๐‘ก ๐‘‘๐‘ก/๐‘‘๐‘ฅ 1+tan^2 ๐‘ฅ=2๐‘ก . ๐‘‘๐‘ก/๐‘‘๐‘ฅ 1+(๐‘ก^2 )^2=2๐‘ก . ๐‘‘๐‘ก/๐‘‘๐‘ฅ 1+๐‘ก^4=2๐‘ก . ๐‘‘๐‘ก/๐‘‘๐‘ฅ (1+๐‘ก^4 ) ๐‘‘๐‘ฅ=2๐‘ก ๐‘‘๐‘ก ๐‘‘๐‘ฅ=2๐‘ก/(1 + ๐‘ก^4 ) ๐‘‘๐‘ก Putting values of t & dt, we get ("Using" ๐‘ก๐‘Ž๐‘›โก๐‘ฅ= ๐‘ก^2) Putting values of t & dt, we get 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€–โˆš(tanโก๐‘ฅ ) (cotโก๐‘ฅ+1) ใ€— ๐‘‘๐‘ฅ =1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’[โˆš(๐‘ก^2 ) (cotโก๐‘ฅ+1)] ๐‘‘๐‘ฅ = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’[โˆš(๐‘ก^2 ) (1/tanโก๐‘ฅ +1)] ๐‘‘๐‘ฅ = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’๐‘ก[1/๐‘ก^2 +1] ๐‘‘๐‘ฅ = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’๐‘ก[(1 + ๐‘ก^2)/๐‘ก^2 ] ๐‘‘๐‘ฅ = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’๐‘ก[(1 + ๐‘ก^2)/๐‘ก^2 ] ร—2๐‘ก/(1 + ๐‘ก^2 ) . ๐‘‘๐‘ก = 1/โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’2[(1 + ๐‘ก^2)/(1 + ๐‘ก^4 )] ๐‘‘๐‘ก = 1/โˆš2 2โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’(1 + ๐‘ก^2)/(1 + ๐‘ก^4 ) ๐‘‘๐‘ก Dividing numerator and denominator by ๐‘ก^2 = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– ((1 + ๐‘ก^2)/๐‘ก^2 )/((1 + ๐‘ก^4)/๐‘ก^2 )ใ€—. ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (1/๐‘ก^2 + 1)/(1/๐‘ก^2 + ๐‘ก^2 )ใ€—. ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’(1 + 1/๐‘ก^2 )/( ๐‘ก^2 + 1/๐‘ก^2 + 2 โˆ’ 2). ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’(1 + 1/๐‘ก^2 )/( (๐‘ก)^2 + (1/๐‘ก)^2โˆ’ 2 (๐‘ก) (1/๐‘ก) + 2). ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (1 + 1/๐‘ก^2 )/((๐‘ก โˆ’ 1/๐‘ก)^2 + 2)ใ€—. ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’(1 + 1/๐‘ก^2 )/((๐‘ก โˆ’ 1/๐‘ก)^2 +(โˆš2 )^2 ). ๐‘‘๐‘ก (Adding and subtracting 2 in denominator) Let ๐‘กโˆ’1/๐‘ก=๐‘ฆ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. 1+ 1/๐‘ก^2 = ๐‘‘๐‘ฆ/๐‘‘๐‘ก ๐‘‘๐‘ก =๐‘‘๐‘ฆ/((1 + 1/๐‘ก^2 ) ) Putting the values of (1/t โˆ’t) and dt, we get โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’(1 + 1/๐‘ก^2 )/((๐‘ก โˆ’ 1/๐‘ก)^2 +(โˆš2 )^2 ). ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– (1 + 1/๐‘ก^2 )/(๐‘ฆ^2 +(โˆš2 )^2 )ใ€—. ๐‘‘๐‘ก = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– ((1 + 1/๐‘ก^2 ))/(๐‘ฆ^2 +(โˆš2 )^2 )ใ€—ร— ๐‘‘๐‘ฆ/((1 โˆ’ 1/๐‘ก^2 ) ) = โˆš2 โˆซ1_(๐œ‹/6)^(๐œ‹/3)โ–’ใ€– 1/(๐‘ฆ^2 +(โˆš2 )^2 )ใ€—. ๐‘‘๐‘ฆ = โˆš2 (1/โˆš2 tan^(โˆ’1)โกใ€– ๐‘ฆ/โˆš2ใ€— )_(๐œ‹/6)^(๐œ‹/3) = (tan^(โˆ’1)โกใ€– ๐‘ฆ/โˆš2ใ€— )_(๐œ‹/6)^(๐œ‹/3) = (tan^(โˆ’1)โกใ€– (1/๐‘ก โˆ’ ๐‘ก)/โˆš2ใ€— )_(๐œ‹/6)^(๐œ‹/3) ["Using" ๐‘ฆ=1/๐‘ก โˆ’๐‘ก] = (tan^(โˆ’1)โกใ€– (๐‘ก^2 โˆ’ 1)/(โˆš2 ๐‘ก)ใ€— )_(๐œ‹/6)^(๐œ‹/3) = (tan^(โˆ’1)โก((tanโก๐‘ฅ โˆ’ 1)/(โˆš2 โˆš(tanโก๐‘ฅ ))) )_(๐œ‹/6)^(๐œ‹/3) = tan^(โˆ’1) ((tanโก(๐œ‹/3) โˆ’ 1)/โˆš(2 tanโก(๐œ‹/3) ))โˆ’tan^(โˆ’1) ((tanโก(๐œ‹/6) โˆ’ 1)/โˆš(2 tanโก(๐œ‹/6) )) = tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) )โˆ’tan^(โˆ’1) ((1/โˆš3 โˆ’1 )/(โˆš3 โˆš(2 . 1/(โˆš3 " " )))) = tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) )โˆ’tan^(โˆ’1) ((1 โˆ’ โˆš3 )/(โˆš3 โˆš(2 . 1/(โˆš3 " " )))) Using ๐‘ก๐‘Ž๐‘›โก(๐œ‹/3) = โˆš3 ๐‘ก๐‘Ž๐‘›โก(๐œ‹/6) = 1/(โˆš3 " " ) = tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) )โˆ’tan^(โˆ’1) ((1 โˆ’ โˆš3 )/โˆš(3 . 2 . 1/(โˆš3 " " ))) = tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) )โˆ’tan^(โˆ’1) ((1 โˆ’ โˆš3 )/โˆš(2 โˆš3) ) = tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) )โˆ’tan^(โˆ’1) ((โˆ’(โˆš3 โˆ’ 1))/โˆš(2 โˆš3) ) = tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) )+tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) ) = 2 tan^(โˆ’1) ((โˆš3 โˆ’ 1)/โˆš(2 โˆš3) ) = ๐Ÿ ใ€–๐ฌ๐ข๐’ใ€—^(โˆ’๐Ÿ) [(โˆš๐Ÿ‘ โˆ’ ๐Ÿ)/๐Ÿ] "Using " tan^(โˆ’1)โก(โˆ’๐œƒ) =โˆ’tan^(โˆ’1)โก(๐œƒ) Note ๐ด๐ต^2=๐ต๐ถ^2+๐ด๐ถ^2 ๐ด๐ต^2=(โˆš3โˆ’1)^2+(โˆš(2 โˆš3) )^2 ๐ด๐ต^2=3+1โˆ’2 โˆš3+2 โˆš3 ๐ด๐ต^2=4 โˆด ๐ด๐ต=2 sin ๐œƒ = ๐ต๐ถ/๐ด๐ต sin ๐œƒ = (โˆš3 โˆ’ 1)/2 ๐œƒ = ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((โˆš3 โˆ’ 1)/2)

Ncert Solution Class 12 Maths Chapter 7 Miscellaneous Exercise

Source: https://www.teachoo.com/4854/728/Misc-28---Chapter-7-Class-12-Integration---Evaluate-definite/category/Miscellaneous/