Integral of cotx
Integral of cotx

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Last updated at May 29, 2023 by Teachoo

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

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