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 Teachoo
Transcript
Misc 28 Evaluate the definite integral ∫_(𝜋/6)^(𝜋/3)▒〖(sin𝑥 + cos𝑥)/√(sin〖2𝑥 〗 ) 〗 ∫_(𝜋/6)^(𝜋/3)▒〖(sin𝑥 + cos𝑥)/√(sin〖2𝑥 〗 ) 𝑑𝑥 〗 = ∫_(𝜋/6)^(𝜋/3)▒〖 (sin𝑥 + cos𝑥)/√(2 sin𝑥 cos𝑥 ) 𝑑𝑥 〗 = ∫_(𝜋/6)^(𝜋/3)▒〖 (sin𝑥/√(2 sin𝑥 cos𝑥 )+cos𝑥/√(2 sin𝑥 cos𝑥 )) 𝑑𝑥 〗 = ∫_(𝜋/6)^(𝜋/3)▒〖 (√(sin𝑥 )/(√2 √(cos𝑥 ))+√(cos𝑥 )/(√2 √(sin𝑥 ))) 𝑑𝑥 〗 ("Using" sin2𝜃=2 sin𝜃 cos𝜃 ) = ∫_(𝜋/6)^(𝜋/3)▒〖 (1/√2 √(sin𝑥/cos𝑥 ) + 1/√2 √(cos𝑥/sin𝑥 )) 𝑑𝑥 〗 = 1/√2 ∫_(𝜋/6)^(𝜋/3)▒〖(√(tan𝑥 )+√(cot𝑥 ) ) 𝑑𝑥 〗 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒〖 [√(cot𝑥 )+1/√(cot𝑥 )] 〗 𝑑𝑥 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒(cot𝑥 + 1)/√(cot𝑥 ) 𝑑𝑥 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒〖√(tan𝑥 ) (cot𝑥+1) 〗 𝑑𝑥 ("Using" 𝑡𝑎𝑛𝑥= 1/𝑐𝑜𝑡𝑥 ) Let tan𝑥=𝑡^2 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. sec^2 𝑥=2𝑡 𝑑𝑡/𝑑𝑥 1+tan^2 𝑥=2𝑡 . 𝑑𝑡/𝑑𝑥 1+(𝑡^2 )^2=2𝑡 . 𝑑𝑡/𝑑𝑥 1+𝑡^4=2𝑡 . 𝑑𝑡/𝑑𝑥 (1+𝑡^4 ) 𝑑𝑥=2𝑡 𝑑𝑡 𝑑𝑥=2𝑡/(1 + 𝑡^4 ) 𝑑𝑡 Putting values of t & dt, we get ("Using" 𝑡𝑎𝑛𝑥= 𝑡^2) Putting values of t & dt, we get 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒〖√(tan𝑥 ) (cot𝑥+1) 〗 𝑑𝑥 =1/√2 ∫1_(𝜋/6)^(𝜋/3)▒[√(𝑡^2 ) (cot𝑥+1)] 𝑑𝑥 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒[√(𝑡^2 ) (1/tan𝑥 +1)] 𝑑𝑥 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒𝑡[1/𝑡^2 +1] 𝑑𝑥 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒𝑡[(1 + 𝑡^2)/𝑡^2 ] 𝑑𝑥 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒𝑡[(1 + 𝑡^2)/𝑡^2 ] ×2𝑡/(1 + 𝑡^2 ) . 𝑑𝑡 = 1/√2 ∫1_(𝜋/6)^(𝜋/3)▒2[(1 + 𝑡^2)/(1 + 𝑡^4 )] 𝑑𝑡 = 1/√2 2∫1_(𝜋/6)^(𝜋/3)▒(1 + 𝑡^2)/(1 + 𝑡^4 ) 𝑑𝑡 Dividing numerator and denominator by 𝑡^2 = √2 ∫1_(𝜋/6)^(𝜋/3)▒〖 ((1 + 𝑡^2)/𝑡^2 )/((1 + 𝑡^4)/𝑡^2 )〗. 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒〖 (1/𝑡^2 + 1)/(1/𝑡^2 + 𝑡^2 )〗. 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒(1 + 1/𝑡^2 )/( 𝑡^2 + 1/𝑡^2 + 2 − 2). 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒(1 + 1/𝑡^2 )/( (𝑡)^2 + (1/𝑡)^2− 2 (𝑡) (1/𝑡) + 2). 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒〖 (1 + 1/𝑡^2 )/((𝑡 − 1/𝑡)^2 + 2)〗. 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒(1 + 1/𝑡^2 )/((𝑡 − 1/𝑡)^2 +(√2 )^2 ). 𝑑𝑡 (Adding and subtracting 2 in denominator) Let 𝑡−1/𝑡=𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 1+ 1/𝑡^2 = 𝑑𝑦/𝑑𝑡 𝑑𝑡 =𝑑𝑦/((1 + 1/𝑡^2 ) ) Putting the values of (1/t −t) and dt, we get √2 ∫1_(𝜋/6)^(𝜋/3)▒(1 + 1/𝑡^2 )/((𝑡 − 1/𝑡)^2 +(√2 )^2 ). 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒〖 (1 + 1/𝑡^2 )/(𝑦^2 +(√2 )^2 )〗. 𝑑𝑡 = √2 ∫1_(𝜋/6)^(𝜋/3)▒〖 ((1 + 1/𝑡^2 ))/(𝑦^2 +(√2 )^2 )〗× 𝑑𝑦/((1 − 1/𝑡^2 ) ) = √2 ∫1_(𝜋/6)^(𝜋/3)▒〖 1/(𝑦^2 +(√2 )^2 )〗. 𝑑𝑦 = √2 (1/√2 tan^(−1)〖 𝑦/√2〗 )_(𝜋/6)^(𝜋/3) = (tan^(−1)〖 𝑦/√2〗 )_(𝜋/6)^(𝜋/3) = (tan^(−1)〖 (1/𝑡 − 𝑡)/√2〗 )_(𝜋/6)^(𝜋/3) ["Using" 𝑦=1/𝑡 −𝑡] = (tan^(−1)〖 (𝑡^2 − 1)/(√2 𝑡)〗 )_(𝜋/6)^(𝜋/3) = (tan^(−1)((tan𝑥 − 1)/(√2 √(tan𝑥 ))) )_(𝜋/6)^(𝜋/3) = tan^(−1) ((tan(𝜋/3) − 1)/√(2 tan(𝜋/3) ))−tan^(−1) ((tan(𝜋/6) − 1)/√(2 tan(𝜋/6) )) = tan^(−1) ((√3 − 1)/√(2 √3) )−tan^(−1) ((1/√3 −1 )/(√3 √(2 . 1/(√3 " " )))) = tan^(−1) ((√3 − 1)/√(2 √3) )−tan^(−1) ((1 − √3 )/(√3 √(2 . 1/(√3 " " )))) Using 𝑡𝑎𝑛(𝜋/3) = √3 𝑡𝑎𝑛(𝜋/6) = 1/(√3 " " ) = tan^(−1) ((√3 − 1)/√(2 √3) )−tan^(−1) ((1 − √3 )/√(3 . 2 . 1/(√3 " " ))) = tan^(−1) ((√3 − 1)/√(2 √3) )−tan^(−1) ((1 − √3 )/√(2 √3) ) = tan^(−1) ((√3 − 1)/√(2 √3) )−tan^(−1) ((−(√3 − 1))/√(2 √3) ) = tan^(−1) ((√3 − 1)/√(2 √3) )+tan^(−1) ((√3 − 1)/√(2 √3) ) = 2 tan^(−1) ((√3 − 1)/√(2 √3) ) = 𝟐 〖𝐬𝐢𝒏〗^(−𝟏) [(√𝟑 − 𝟏)/𝟐] "Using " tan^(−1)(−𝜃) =−tan^(−1)(𝜃) Note 𝐴𝐵^2=𝐵𝐶^2+𝐴𝐶^2 𝐴𝐵^2=(√3−1)^2+(√(2 √3) )^2 𝐴𝐵^2=3+1−2 √3+2 √3 𝐴𝐵^2=4 ∴ 𝐴𝐵=2 sin 𝜃 = 𝐵𝐶/𝐴𝐵 sin 𝜃 = (√3 − 1)/2 𝜃 = 〖𝑠𝑖𝑛〗^(−1) ((√3 − 1)/2)
Ncert Solution Class 12 Maths Chapter 7 Miscellaneous Exercise
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