Example 6 (iii) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 6 Find the following integrals (iii) ∫1▒1/(1 + tan𝑥 ) 𝑑𝑥 The given function cannot be integrated by direct substitution, so we have to simplify the given function . Simplifying the given function and integrating. ∫1▒1/(1 + tan𝑥 ) .𝑑𝑥 = ∫1▒1/(1 + sin𝑥/cos𝑥 ) .𝑑𝑥 = ∫1▒1/((cos𝑥 + sin𝑥)/cos𝑥 ) .𝑑𝑥 (𝑈𝑠𝑖𝑛𝑔 tan𝑥=sin𝑥/cos𝑥 ) = ∫1▒cos𝑥/(sin𝑥 + cos𝑥 ) .𝑑𝑥 Multiplying and dividing by 2 = ∫1▒(2 cos𝑥)/(2 (sin𝑥 + cos𝑥 ) ) .𝑑𝑥 = ∫1▒(cos𝑥 + cos𝑥)/(2 (sin𝑥 + cos𝑥 ) ) .𝑑𝑥 Adding and subtracting sin𝑥 in the numerator = ∫1▒(cos𝑥 + cos𝑥 + sin𝑥 − sin𝑥)/(2 (sin𝑥 + cos𝑥 ) ) .𝑑𝑥 = 1/2 ∫1▒(sin𝑥 + cos𝑥 + cos𝑥 − sin𝑥)/(sin𝑥 + cos𝑥 ) .𝑑𝑥 = 1/2 ∫1▒[(sin𝑥 + cos𝑥)/(sin𝑥 + cos𝑥 )+〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 )] 𝑑𝑥 = 1/2 ∫1▒[1+〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 )] 𝑑𝑥 = 1/2 [∫1▒〖1.𝑑𝑥+∫1▒〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 )〗 𝑑𝑥] = 1/2 [𝑥+𝐶1+∫1▒〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 ) 𝑑𝑥] Take, I1 =∫1▒〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 ) .𝑑𝑥 Let sin𝑥 + cos𝑥=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. cos𝑥−sin𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(cos𝑥 − sin𝑥 ) Putting value of (𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 ) and 𝑑𝑥 in I1 . I1=∫1▒〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 ) .𝑑𝑥 I1 = ∫1▒〖cos𝑥 − sin〗𝑥/𝑡 .𝑑𝑡/(cos𝑥 − sin𝑥 ) I1 =∫1▒1/𝑡 .𝑑𝑡 Let sin𝑥 + cos𝑥=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. cos𝑥−sin𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(cos𝑥 − sin𝑥 ) Putting value of (𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 ) and 𝑑𝑥 in I1 . I1=∫1▒〖cos𝑥 − sin〗𝑥/(sin𝑥 + cos𝑥 ) .𝑑𝑥 I1 = ∫1▒〖cos𝑥 − sin〗𝑥/𝑡 .𝑑𝑡/(cos𝑥 − sin𝑥 ) I1 =∫1▒1/𝑡 .𝑑𝑡 (𝑈𝑠𝑖𝑛𝑔 ∫1▒〖1/𝑥 .〗 𝑑𝑥=log〖 |𝑥|〗+𝐶) (𝑈𝑠𝑖𝑛𝑔 𝑡=sin𝑥+cos𝑥 ) (𝑊ℎ𝑒𝑟𝑒 𝐶=𝐶1/2+𝐶2/2)
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo