Ex 7.2, 32 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.2, 32 Integrate 1/(1 + cot⁡𝑥 ) Simplify the given function ∫1▒1/(1 + cot⁡𝑥 ) 𝑑𝑥 = ∫1▒1/(1 + cos⁡𝑥/sin⁡𝑥 ) 𝑑𝑥 = ∫1▒1/(〖sin⁡𝑥 + cos〗⁡𝑥/sin⁡𝑥 ) 𝑑𝑥 = ∫1▒sin⁡𝑥/〖sin⁡𝑥 + cos〗⁡𝑥 𝑑𝑥 Multiplying & dividing by 2 = ∫1▒(2 sin⁡𝑥)/2(〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥 Adding & subtracting 𝑐𝑜𝑠⁡𝑥 in numerator = ∫1▒(sin⁡𝑥 + sin⁡𝑥 + cos⁡𝑥 − cos⁡𝑥)/2(〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒((sin⁡𝑥 + cos⁡𝑥 + sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒((sin⁡𝑥 + cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 +(sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥 = 1/2 ∫1▒(1+(sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥 = 1/2 [𝑥+∫1▒((sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥] + 𝐶1 …(1) Solving 𝐈1 I1 = ∫1▒(sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 𝑑𝑥 Let 〖sin⁡𝑥 + cos〗⁡𝑥=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 〖cos⁡𝑥−sin〗⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/〖cos⁡𝑥 − sin〗⁡𝑥 𝑑𝑥=𝑑𝑡/(−(〖sin⁡𝑥 − cos〗⁡𝑥 ) ) Thus, our equation becomes …(2) I1 = ∫1▒(sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 𝑑𝑥 = ∫1▒(sin⁡𝑥 − cos⁡𝑥)/𝑡 . 𝑑𝑡/(−(〖sin⁡𝑥 − cos〗⁡𝑥 ) ) = −1∫1▒𝑑𝑡/𝑡 = −〖log 〗⁡|𝑡|+𝐶 Putting back 𝑡=𝑠𝑖𝑛⁡𝑥+𝑐𝑜𝑠⁡𝑥 = −log⁡〖 |sin⁡𝑥+cos⁡𝑥 |〗+𝐶2 Putting the value of I1 in (1) ∴ ∫1▒〖1/(1 + cot⁡𝑥 ) " " 〗 = 1/2 [𝑥+∫1▒((sin⁡𝑥 − cos⁡𝑥)/〖sin⁡𝑥 + cos〗⁡𝑥 ) 𝑑𝑥] + 𝐶1 = 1/2 [𝑥−log⁡|sin⁡𝑥+cos⁡𝑥 |+𝐶2" " ] +𝐶1 = 𝑥/2−1/2 log⁡〖 |sin⁡𝑥+cos⁡𝑥 |〗+𝐶1+𝐶2/2 = 𝒙/𝟐 −𝟏/𝟐 𝒍𝒐𝒈⁡〖 |𝒔𝒊𝒏⁡𝒙+𝒄𝒐𝒔⁡𝒙 |〗+𝑪