1. Chapter 7 Class 12 Integrals
  2. Serial order wise
  3. Examples
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Example 6 (ii) - Chapter 7 Class 12 Integrals

Last updated at Dec. 16, 2024 by Teachoo

Transcript

Example 6 Find the following integrals (ii) ∫1▒sin⁡𝑥/sin⁡(𝑥 + 𝑎) 𝑑𝑥 Let 𝑥+𝑎=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. 1=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡 Hence, our equation becomes ∫1▒sin⁡𝑥/sin⁡(𝑥 + 𝑎) 𝑑𝑥 Putting the value of (𝑥⁡+ 𝑎) and 𝑑𝑥 = ∫1▒sin⁡𝑥/sin⁡𝑡 𝑑𝑥" " = ∫1▒𝒔𝒊𝒏⁡(𝒕 − 𝒂)/sin⁡𝑡 𝑑𝑡" " = ∫1▒(𝒔𝒊𝒏⁡𝒕 𝐜𝐨𝐬⁡𝒂 − 𝐬𝐢𝐧⁡𝒂 𝐜𝐨𝐬⁡𝒕)/sin⁡𝑡 𝑑𝑡 = ∫1▒(sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 − (sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 ) 𝑑𝑡 = ∫1▒sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 . 𝑑𝑡−∫1▒(sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 . 𝑑𝑡 = ∫1▒cos⁡𝑎 . 𝑑𝑡−∫1▒〖sin⁡𝑎 co𝑡⁡𝑡 〗 . 𝑑𝑡 = cos 𝑎∫1▒1. 𝑑𝑡−sin⁡𝑎 ∫1▒𝒄𝒐𝒕⁡𝒕 . 𝒅𝒕 = cos 𝑎 (𝑡)−sin⁡𝑎 𝒍𝒐𝒈⁡|𝒔𝒊𝒏⁡𝒕 |+ 𝐶 = 𝑡.cos⁡𝑎−𝑠𝑖𝑛⁡𝑎 𝑙𝑜𝑔⁡|𝑠𝑖𝑛⁡𝑡 |+ 𝐶 Putting back value of t = x + a (Using x + a = t ∴ x = t – a) (█(𝑈𝑠𝑖𝑛𝑔 sin⁡(𝑎−𝑏)=@sin⁡𝑎 cos⁡𝑏−sin⁡𝑏 cos⁡𝑎 )) ( ∫1▒co𝑡⁡𝑥 𝑑𝑥=log⁡|sin⁡𝑥 | ) = ∫1▒𝒔𝒊𝒏⁡(𝒕 − 𝒂)/sin⁡𝑡 𝑑𝑡" " = ∫1▒𝒔𝒊𝒏⁡〖𝒕 𝐜𝐨𝐬⁡𝒂 −𝐬𝐢𝐧⁡𝒂 𝐜𝐨𝐬⁡𝒕 〗/sin⁡𝑡 𝑑𝑡 = ∫1▒(sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 − (sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 ) 𝑑𝑡 = ∫1▒sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 . 𝑑𝑡−∫1▒(sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 . 𝑑𝑡 = ∫1▒cos⁡𝑎 . 𝑑𝑡−∫1▒〖sin⁡𝑎 co𝑡⁡𝑡 〗 . 𝑑𝑡 = cos 𝑎∫1▒1. 𝑑𝑡−sin⁡𝑎 ∫1▒𝒄𝒐𝒕⁡𝒕 . 𝒅𝒕 = cos 𝑎 (𝑡)−sin⁡𝑎 [𝒍𝒐𝒈⁡|𝒔𝒊𝒏⁡𝒕 | ]+𝐶1 = 𝑡.cos⁡𝑎−sin⁡𝑎 [log⁡|sin⁡𝑡 | ]+ 𝐶1 Putting back value of t = x + a = (𝑥+𝑎) cos⁡𝑎−sin⁡𝑎 [log⁡|sin⁡(𝑥+𝑎) | ]+ 𝐶1 (Using x + a = t x = t – a) (█(𝑈𝑠𝑖𝑛𝑔 sin⁡(𝑎−𝑏)=@sin⁡𝑎 cos⁡𝑏−sin⁡𝑏 cos⁡𝑎 )) ( ∫1▒co𝑡⁡𝑥 𝑑𝑥=log⁡|sin⁡𝑥 | ) = (𝑥+𝑎) cos⁡𝑎−sin⁡𝑎 log⁡〖 |sin⁡(𝑥+𝑎) |〗+ 𝐶 = 𝑥 cos⁡𝑎+〖𝑎 cos〗⁡𝑎−sin⁡𝑎 log⁡|sin⁡(𝑥+𝑎) |+𝐶 = 𝑥 cos⁡𝑎−sin⁡𝑎 log⁡|sin⁡(𝑥+𝑎) |+〖𝒂 𝒄𝒐𝒔〗⁡𝒂+𝑪 = 𝒙 𝒄𝒐𝒔⁡𝒂−𝐬𝐢𝐧⁡〖 𝒂〗 𝒍𝒐𝒈⁡|𝒔𝒊𝒏⁡(𝒙+𝒂) |+𝐂𝟏 (𝑊ℎ𝑒𝑟𝑒 𝐶1=𝑎 cos⁡𝑎+𝐶)
  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

    3. Examples

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Chapter 7 Class 12 Integrals
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Davneet Singh

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