Ex 7.2, 33 - Chapter 7 Class 12 Integrals (Term 2)

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Ex 7.2, 33 Integrate 1/(1 − 𝑡𝑎𝑛⁡𝑥 ) ∫1▒1/(1 − tan⁡𝑥 ) 𝑑𝑥 = ∫1▒1/(1 − sin⁡𝑥/cos⁡𝑥 ) 𝑑𝑥 = ∫1▒1/(〖cos⁡𝑥 − sin〗⁡𝑥/cos⁡𝑥 ) 𝑑𝑥 = ∫1▒cos⁡𝑥/〖cos⁡𝑥 − sin〗⁡𝑥 𝑑𝑥 = ∫1▒(2 cos⁡𝑥)/2(〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = ∫1▒(cos⁡𝑥 + cos⁡𝑥)/2(〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = ∫1▒(cos⁡𝑥 + cos⁡𝑥 + sin⁡𝑥 − sin⁡𝑥)/2(〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = ∫1▒(cos⁡𝑥 − sin⁡𝑥 + cos⁡𝑥 + sin⁡𝑥)/2(〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒((cos⁡𝑥 − sin⁡𝑥 + cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒((cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 +(cos⁡𝑥 − sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒(1+ (cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒𝑑𝑥 + 1/2 ∫1▒(cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 𝑑𝑥 (Adding & subtracting 𝑠𝑖𝑛⁡𝑥 in numerator) = 𝑥/2+1/2 ∫1▒(cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 𝑑𝑥+𝐶_1 Solving 𝐈1 I1 = ∫1▒(cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 .𝑑𝑥 Let cos⁡𝑥 − sin⁡𝑥=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 〖〖 −𝑠𝑖𝑛〗⁡𝑥−𝑐𝑜𝑠〗⁡𝑥=𝑑𝑡/𝑑𝑥 −(〖𝑠𝑖𝑛⁡𝑥+𝑐𝑜𝑠〗⁡𝑥 )=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(−(〖sin⁡𝑥 − cos〗⁡𝑥 ) ) Thus, our equation becomes I1 = ∫1▒(cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 . 𝑑𝑥 I1 = ∫1▒(cos⁡𝑥 − sin⁡𝑥)/𝑡 . 𝑑𝑡/(−(〖sin⁡𝑥 − cos〗⁡𝑥 ) ) I1 = −1∫1▒𝑑𝑡/𝑡 I1 = −log⁡|𝑡|+𝐶2 I1 = −log⁡|cos⁡𝑥−sin⁡𝑥 |+𝐶2 (Using ∫1▒1/𝑥. 𝑑𝑥=𝑙𝑜𝑔⁡|𝑥|) (Using 〖〖𝑡=𝑐𝑜𝑠〗⁡𝑥−𝑠𝑖𝑛〗⁡𝑥 ) Putting the value of I1 in (1) ∴ ∫1▒〖1/(1 + tan⁡𝑥 ) " " 〗= 𝑥/2+1/2 ∫1▒(cos⁡𝑥 + sin⁡𝑥)/〖cos⁡𝑥 − sin〗⁡𝑥 𝑑𝑥+𝐶_1 = 𝑥/2+1/2(−𝐥𝐨𝐠⁡〖 |𝒄𝒐𝒔⁡𝒙−𝒔𝒊𝒏⁡𝒙 |〗)+𝐶2/2+𝐶1 = 𝒙/𝟐 −𝟏/𝟐 𝒍𝒐𝒈⁡〖 |𝒄𝒐𝒔⁡𝒙−𝒔𝒊𝒏⁡𝒙 |〗+𝑪