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- 1. ANSWERS TO QUESTIONSFOR REVISION CLASS XII : MATHEMATICS TOPICS: INTEGRAL CALCULUS PREPARED BY M SRINIVASAN, PGT(MATHS), KVS
- 2. 𝟏. 𝒆𝒙 𝒄𝒐𝒔𝒙 − 𝒔𝒊𝒏 𝒙 𝒅𝒙 is equal to 𝒂) 𝒆𝒙 𝒄𝒐𝒔 𝒙 + C b) 𝒆𝒙 𝒔𝒊𝒏 𝒙 + C c) −𝒆𝒙 𝒄𝒐𝒔 𝒙 + 𝑪 d) −𝒆𝒙 𝒔𝒊𝒏 𝒙 + 𝑪 𝒆𝒙 𝒄𝒐𝒔 𝒙 − 𝒔𝒊𝒏 𝒙 𝒅𝒙 𝒆𝒙 𝒇 𝒙 + 𝒇′(𝒙) 𝒅𝒙 = 𝒆𝒙 𝒇 𝒙 + 𝑪 𝒆𝒙 𝒄𝒐𝒔 𝒙 − 𝒔𝒊𝒏 𝒙 𝒅𝒙 = 𝒆𝒙 𝒄𝒐𝒔 𝒙 + (− 𝒔𝒊𝒏 𝒙 𝒅𝒙 𝒇 𝒙 = 𝐜𝐨𝐬 𝒙 𝒇′ 𝒙 = −𝐬𝐢𝐧 𝒙 න 𝒆𝒙 𝒄𝒐𝒔 𝒙 + (− 𝒔𝒊𝒏 𝒙 𝒅𝒙 = 𝒆𝒙 𝒄𝒐𝒔 𝒙 + C 𝒂) 𝒆𝒙 𝒄𝒐𝒔 𝒙 + C
- 3. 2. −𝟏 𝟏 𝒙−𝟐 𝒙−𝟐 𝒅𝒙,𝒙 ≠ 𝟎 is equal to a)1 b −𝟏 c) 2 d) −𝟐 න −𝟏 𝟏 𝒙 − 𝟐 𝒙 − 𝟐 𝒅𝒙 𝒙 − 𝟐 = ቊ 𝒙 − 𝟐 , 𝒙 − 𝟐 ≥ 𝟎 − 𝒙 − 𝟐 , 𝒙 − 𝟐 < 𝟎 𝒙 − 𝟐 = ቊ 𝒙 − 𝟐 , 𝒙 ≥ 𝟐 − 𝒙 − 𝟐 , 𝒙 < 𝟐 න −𝟏 𝟏 𝒙 − 𝟐 𝒙 − 𝟐 𝒅𝒙 = න −𝟏 𝟏 −(𝒙 − 𝟐) 𝒙 − 𝟐 𝒅𝒙 = − න −𝟏 𝟏 𝒅𝒙 = − 𝑥 −1 1 = − 1 − (−1) = −2 𝒅) − 𝟐
- 4. 3. The valueof 𝑺𝒊𝒏−𝟏 𝒄𝒐𝒔 𝝅 𝟐 − 𝟑𝒙 𝒅𝒙 is equal to a) 𝒙𝟐 𝟐 + 𝑪 b) 𝟑𝒙𝟐 𝟐 + 𝑪 c) 𝑺𝒊𝒏 (𝟑𝒙𝟐) 𝟐 + 𝑪 d) 𝑪𝒐𝒔 (𝟑𝒙𝟐) 𝟐 + 𝑪 න 𝑺𝒊𝒏−𝟏 𝒄𝒐𝒔 𝝅 𝟐 − 𝟑𝒙 𝒅𝒙 = න 𝑺𝒊𝒏−𝟏 𝒔𝒊𝒏 𝟑𝒙 𝒅𝒙 = න 𝟑𝒙 𝒅𝒙 = 𝟑𝒙𝟐 𝟐 + 𝑪 b) 𝟑𝒙𝟐 𝟐 + 𝑪
- 5. 4. 𝟐 𝒆𝒙+𝒆−𝒙 𝟐𝒅𝒙 is equal to a) − 𝒆𝒙 𝒆𝒙+𝒆−𝒙 + 𝑪 b) 𝟏 𝒆𝒙+𝒆−𝒙 + 𝑪 c) 𝟏 𝒆𝒙+𝟏 𝟐 + 𝑪 d) − 𝟏 𝒆𝒙+𝒆−𝒙 + 𝑪 න 𝟐 𝒆𝒙 + 𝒆−𝒙 𝟐 𝒅𝒙 = න 𝟐 𝒆𝒙 + 𝟏 𝒆𝒙 𝟐 𝒅𝒙 𝟐 𝒆𝒙+𝒆−𝒙 𝟐 𝒅𝒙 = 𝟐𝒆𝟐𝒙 𝒆𝟐𝒙+𝟏 𝟐 𝒅𝒙
- 6. 𝒖 = 𝒆𝟐𝒙 +𝟏 𝒅𝒖 = 𝟐 𝒆𝟐𝒙 𝒅𝒙 න 𝟐𝒆𝟐𝒙 𝒆𝟐𝒙 + 𝟏 𝟐 𝒅𝒙 = න 𝒅𝒖 𝒖 𝟐 𝒅𝒙 𝟐 𝒆𝒙+𝒆−𝒙 𝟐 𝒅𝒙 = 𝟐𝒆𝟐𝒙 𝒆𝟐𝒙+𝟏 𝟐 𝒅𝒙 = − 𝟏 𝒖 + 𝑪 = − 𝟏 𝒆𝟐𝒙 + 𝟏 + 𝑪 = − 𝒆−𝒙 𝒆𝒙 + 𝒆−𝒙 + 𝑪 a) − 𝒆𝒙 𝒆𝒙 + 𝒆−𝒙 + 𝑪 Multiply and divide by 𝒆−𝒙
- 7. 5. The valueof is 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒+𝟑 𝒔𝒊𝒏 𝒙 𝟒+𝟑 𝒄𝒐𝒔 𝒙 𝒅𝒙 a) 2 b) 𝟑 𝟒 c) 0 d) −𝟐 න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒 + 𝟑 𝒔𝒊𝒏 𝒙 𝟒 + 𝟑 𝒄𝒐𝒔 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒 + 𝟑 𝒔𝒊𝒏 𝝅 𝟐 − 𝒙 𝟒 + 𝟑 𝒄𝒐𝒔 𝝅 𝟐 − 𝒙 𝒅𝒙 න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒 + 𝟑 𝒔𝒊𝒏 𝒙 𝟒 + 𝟑 𝒄𝒐𝒔 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒 + 𝟑 𝒄𝒐𝒔 𝒙 𝟒 + 𝟑 𝒔𝒊𝒏 𝒙 𝒅𝒙 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒+𝟑 𝒔𝒊𝒏 𝒙 𝟒+𝟑 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝟏 𝟐 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒+𝟑 𝒔𝒊𝒏 𝒙 𝟒+𝟑 𝒄𝒐𝒔 𝒙 + 𝒍𝒐𝒈 𝟒+𝟑 𝒄𝒐𝒔 𝒙 𝟒+𝟑 𝒔𝒊𝒏 𝒙 𝒅𝒙 = 𝟏 𝟐 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟒+𝟑 𝒔𝒊𝒏 𝒙 𝟒+𝟑 𝒄𝒐𝒔 𝒙 × 𝟒+𝟑 𝒄𝒐𝒔 𝒙 𝟒+𝟑 𝒔𝒊𝒏 𝒙 𝒅𝒙 = 𝟏 𝟐 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟏 𝒅𝒙 = 𝟏 𝟐 𝟎 𝝅 𝟐 𝟎 𝒅𝒙 c) 𝟎
- 8. 6. Assertion : 𝟐 𝟖 𝟏𝟎−𝒙 𝒙+ 𝟏𝟎−𝒙 𝒅𝒙 = 𝟑 Reason : 𝒂 𝒃 𝒇 𝒙 𝒅𝒙 = 𝒂 𝒃 𝒇(𝒂 + 𝒃 − 𝒙) 𝒅𝒙 Assertion න 𝟐 𝟖 𝟏𝟎 − 𝒙 𝒙 + 𝟏𝟎 − 𝒙 𝒅𝒙 = න 𝟐 𝟖 𝟏𝟎 − 𝟐 + 𝟖 − 𝒙 𝟐 + 𝟖 − 𝒙 + 𝟏𝟎 − 𝟐 + 𝟖 − 𝒙 𝒅𝒙 = න 𝟐 𝟖 𝟏𝟎 − 𝟏𝟎 − 𝒙 𝟏𝟎 − 𝒙 + 𝟏𝟎 − 𝟏𝟎 − 𝒙 𝒅𝒙 න 𝟐 𝟖 𝟏𝟎 − 𝒙 𝒙 + 𝟏𝟎 − 𝒙 𝒅𝒙 = න 𝟐 𝟖 𝒙 𝟏𝟎 − 𝒙 + 𝒙 𝒅𝒙
- 9. න 𝟐 𝟖 𝟏𝟎 − 𝒙 𝒙+ 𝟏𝟎 − 𝒙 𝒅𝒙 = න 𝟐 𝟖 𝒙 𝟏𝟎 − 𝒙 + 𝒙 𝒅𝒙 න 𝟐 𝟖 𝟏𝟎 − 𝒙 𝒙 + 𝟏𝟎 − 𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝟐 𝟖 𝟏𝟎 − 𝒙 𝒙 + 𝟏𝟎 − 𝒙 + 𝒙 𝟏𝟎 − 𝒙 + 𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝟐 𝟖 𝟏𝟎 − 𝒙 + 𝒙 𝒙 + 𝟏𝟎 − 𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝟐 𝟖 𝒅𝒙 = 𝟏 𝟐 𝒙 𝟐 𝟖 = 𝟏 𝟐 𝟖 − 𝟐 = 𝟏 𝟐 𝟔 = 𝟑
- 10. න 𝟐 𝟖 𝟏𝟎 − 𝒙 𝒙+ 𝟏𝟎 − 𝒙 𝒅𝒙 = 𝟑 Assertion is True Reason න 𝒂 𝒃 𝒇 𝒙 𝒅𝒙 = න 𝒂 𝒃 𝒇(𝒂 + 𝒃 − 𝒙) 𝒅𝒙 Reason is True Reason is the correct explanation of Assertion a) Both A and R are True and R is the correct explanation of A
- 11. 7. Assertion :If 𝒇′ 𝒙 = 𝒙 + 𝟏 𝟏+𝒙𝟐 and 𝒇(𝟎) = 𝟎 then 𝒇 𝒙 = 𝒙𝟐 𝟐 + 𝒕𝒂𝒏−𝟏 𝒙 Reason : 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝐥𝐨𝐠(𝒔𝒆𝒄 𝒙) + 𝑪. Assertion 𝒇′ 𝒙 = 𝒙 + 𝟏 𝟏 + 𝒙𝟐 ⟹ 𝒇 𝒙 = න 𝒇′(𝒙) 𝒅𝒙 𝒇 𝒙 = න 𝒙 + 𝟏 𝟏 + 𝒙𝟐 𝒅𝒙 𝒇 𝒙 = 𝒙𝟐 𝟐 + 𝒕𝒂𝒏−𝟏 𝒙 + 𝑪 𝒇 𝟎 = 𝟎 ⟹ 𝒇 𝟎 = 𝟎 𝟐 + 𝒕𝒂𝒏−𝟏 𝟎 + 𝑪 = 𝟎 𝑪 = 𝟎 𝒇 𝒙 = 𝒙𝟐 𝟐 + 𝒕𝒂𝒏−𝟏 𝒙 Assertion is True
- 12. Reason Reason is TrueReason is not used to prove Assertion න 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝐥𝐨𝐠(𝒔𝒆𝒄 𝒙) + 𝑪 b) Both A and R are True and R is not the correct explanation of A
- 13. 8. Evaluate: 𝒙𝟗 𝟒𝒙𝟐+𝟏 𝟔𝒅𝒙 න 𝒙𝟗 𝟒𝒙𝟐 + 𝟏 𝟔 = න 𝒙𝟗 𝒙𝟐 𝟔 𝟒 + 𝟏 𝒙𝟐 𝟔 𝒅𝒙 = න 𝒙𝟗 𝒙𝟏𝟐 𝟒 + 𝟏 𝒙𝟐 𝟔 𝒅𝒙 න 𝒙𝟗 𝟒𝒙𝟐 + 𝟏 𝟔 = න 𝟏 𝒙𝟑 𝟒 + 𝟏 𝒙𝟐 𝟔
- 14. න 𝒙𝟗 𝟒𝒙𝟐 + 𝟏𝟔 𝒅𝒙 = න 𝒅𝒙 𝒙𝟑 𝟒 + 𝟏 𝒙𝟐 𝟔 𝒖 = 𝟏 𝒙𝟐 𝒅𝒖 = −𝟐 𝒙𝟑 𝒅𝒙 𝒅𝒙 𝒙𝟑 = − 𝟏 𝟐 𝒅𝒖 න 𝒅𝒙 𝒙𝟑 𝟒 + 𝟏 𝒙𝟐 𝟔 = න − 𝟏 𝟐 𝒅𝒖 𝟒 + 𝒖 𝟔 = − 𝟏 𝟐 𝟒 + 𝒖 −𝟓 −𝟓 = − 𝟏 𝟐 න 𝟒 + 𝒖 −𝟔𝒅𝒖 න 𝒙𝟗 𝟒𝒙𝟐 + 𝟏 𝟔 𝒅𝒙 = 𝟏 𝟏𝟎 𝟒 + 𝟏 𝒙𝟐 −𝟓 + 𝑪
- 15. 9. Evaluate: 𝒔𝒊𝒏𝒙−𝒙 𝒄𝒐𝒔 𝒙 𝒙 (𝒙+𝒔𝒊𝒏 𝒙) 𝒅𝒙 න 𝒔𝒊𝒏 𝒙 − 𝒙 𝒄𝒐𝒔 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = න 𝒙 + 𝒔𝒊𝒏 𝒙 − 𝒙 − 𝒙 𝒄𝒐𝒔 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = න 𝒙 + 𝒔𝒊𝒏 𝒙 − 𝒙 𝟏 + 𝒄𝒐𝒔 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = න 𝒙 + 𝒔𝒊𝒏 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) − 𝒙 (𝟏 + 𝒄𝒐𝒔 𝒙) 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = න 𝟏 𝒙 − (𝟏 + 𝒄𝒐𝒔 𝒙) (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = න 𝟏 𝒙 𝒅𝒙 − න (𝟏 + 𝒄𝒐𝒔 𝒙) (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙
- 16. න 𝒔𝒊𝒏 𝒙 −𝒙 𝒄𝒐𝒔 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = න 𝟏 𝒙 𝒅𝒙 − න (𝟏 + 𝒄𝒐𝒔 𝒙) (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 𝑳𝒆𝒕 𝒖 = 𝒙 + 𝐬𝐢𝐧 𝒙 𝒅𝒖 = 𝟏 + 𝒄𝒐𝒔 𝒙 𝒅𝒙 න 𝒔𝒊𝒏 𝒙 − 𝒙 𝒄𝒐𝒔 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = 𝒍𝒐𝒈 𝒙 − න 𝒅𝒖 𝒖 = 𝒍𝒐𝒈 𝒙 − 𝒍𝒐𝒈 𝒖 + 𝑪 = 𝒍𝒐𝒈 𝒙 − 𝒍𝒐𝒈 (𝒙 + 𝒔𝒊𝒏 𝒙) + 𝑪 න 𝒔𝒊𝒏 𝒙 − 𝒙 𝒄𝒐𝒔 𝒙 𝒙 (𝒙 + 𝒔𝒊𝒏 𝒙) 𝒅𝒙 = 𝒍𝒐𝒈 𝒙 − 𝒍𝒐𝒈 (𝒙 + 𝒔𝒊𝒏 𝒙) + 𝑪
- 17. 10. Evaluate: 𝒔𝒊𝒏 𝟐𝒙 𝒕𝒂𝒏−𝟏 𝒔𝒊𝒏 𝒙 𝒅𝒙 න 𝒔𝒊𝒏 𝟐𝒙 𝒕𝒂𝒏−𝟏 𝒔𝒊𝒏 𝒙 𝒅𝒙 = න 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒕𝒂𝒏−𝟏 (𝒔𝒊𝒏 𝒙) 𝒅𝒙 = 𝟐 න 𝒔𝒊𝒏 𝒙 𝒕𝒂𝒏−𝟏(𝒔𝒊𝒏 𝒙) 𝒄𝒐𝒔 𝒙𝒅𝒙 𝒕 = 𝒔𝒊𝒏 𝒙 𝒅𝒕 = 𝒄𝒐𝒔 𝒙 𝒅𝒙 𝟐 න 𝒔𝒊𝒏 𝒙 𝒕𝒂𝒏−𝟏(𝒔𝒊𝒏 𝒙) 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝟐 න 𝒕 𝒕𝒂𝒏−𝟏 𝒕 𝒅𝒕 Let 𝒖 = 𝒕𝒂𝒏−𝟏 𝒕 𝒅𝒗 = 𝒕 𝒅𝒕 𝒅𝒖 = 𝒅𝒕 𝟏 + 𝒕𝟐 𝒗 = 𝒕𝟐 𝟐 𝟐 න 𝒕 𝒕𝒂𝒏−𝟏 𝒕 𝒅𝒕 = 𝟐 𝒕𝟐 𝟐 𝒕𝒂𝒏−𝟏 𝒕 − න 𝒕𝟐 𝟐 𝒅𝒕 𝟏 + 𝒕𝟐
- 18. 𝟐 න 𝒕𝒕𝒂𝒏−𝟏 𝒕 𝒅𝒕 = 𝟐 𝒕𝟐 𝟐 𝒕𝒂𝒏−𝟏 𝒕 − න 𝒕𝟐 𝟐 𝒅𝒕 𝟏 + 𝒕𝟐 = 𝒕𝟐𝒕𝒂𝒏−𝟏 𝒕 − න 𝒕𝟐 𝟏 + 𝒕𝟐 𝒅𝒕 = 𝒕𝟐𝒕𝒂𝒏−𝟏 𝒕 − න 𝒕𝟐 + 𝟏 − 𝟏 𝟏 + 𝒕𝟐 𝒅𝒕 = 𝒕𝟐 𝒕𝒂𝒏−𝟏 𝒕 − න 𝒕𝟐 + 𝟏 𝟏 + 𝒕𝟐 − 𝟏 𝟏 + 𝒕𝟐 𝒅𝒕 = 𝒕𝟐𝒕𝒂𝒏−𝟏 𝒕 − න 𝟏 − 𝟏 𝟏 + 𝒕𝟐 𝒅𝒕 = 𝒕𝟐 𝒕𝒂𝒏−𝟏 𝒕 − 𝒕 + 𝒕𝒂𝒏−𝟏 𝒕 + 𝑪 න 𝒔𝒊𝒏 𝟐𝒙 𝒕𝒂𝒏−𝟏 𝒔𝒊𝒏 𝒙 𝒅𝒙 = 𝒔𝒊𝒏𝟐𝒙 𝒕𝒂𝒏−𝟏 𝒔𝒊𝒏𝒙 − 𝒔𝒊𝒏 𝒙 + 𝒕𝒂𝒏−𝟏 𝒔𝒊𝒏 𝒙 + 𝑪
- 19. 11. Evaluate: 𝒆𝒙 𝟏+𝒔𝒊𝒏 𝟐𝒙 𝟏+𝒄𝒐𝒔 𝟐𝒙 𝒅𝒙 න 𝒆𝒙 𝟏 + 𝒔𝒊𝒏 𝟐𝒙 𝟏 + 𝒄𝒐𝒔 𝟐𝒙 𝒅𝒙 = න 𝒆𝒙 𝒔𝒊𝒏𝟐𝒙 + 𝒄𝒐𝒔𝟐𝒙 + 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝟐 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙 = න 𝒆𝒙 𝒔𝒊𝒏 𝒙 + 𝒄𝒐𝒔 𝒙 𝟐 𝟐 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙 = න 𝒆𝒙 𝒔𝒊𝒏 𝒙 + 𝒄𝒐𝒔 𝒙 𝟐 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙
- 20. 𝒆𝒙 𝟏+𝒔𝒊𝒏𝟐𝒙 𝟏+𝒄𝒐𝒔 𝟐𝒙 𝒅𝒙 = 𝒆𝒙 𝒔𝒊𝒏 𝒙+𝒄𝒐𝒔 𝒙 𝟐 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝒆𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝒆𝒙 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 + 𝒔𝒆𝒄 𝒙 𝒅𝒙 𝒆𝒙 𝒇 𝒙 + 𝒇′(𝒙) 𝒅𝒙 = 𝒆𝒙 𝒇 𝒙 + 𝑪 𝒇 𝒙 = 𝐬𝐞𝐜 𝒙 𝒇′ 𝒙 = 𝐬𝐞𝐜 𝒙 𝒕𝒂𝒏 𝒙 𝒆𝒙 𝟏+𝒔𝒊𝒏 𝟐𝒙 𝟏+𝒄𝒐𝒔 𝟐𝒙 𝒅𝒙 = 𝒆𝒙 𝒔𝒆𝒄 𝒙 + 𝑪
- 21. 12. Evaluate: 𝟎 𝝅 𝟐𝟐 𝒍𝒐𝒈(𝒔𝒊𝒏 𝒙) − 𝒍𝒐𝒈(𝒔𝒊𝒏 𝟐𝒙) 𝒅𝒙 න 𝟎 𝝅 𝟐 𝟐 𝒍𝒐𝒈(𝒔𝒊𝒏 𝒙) − 𝒍𝒐𝒈(𝒔𝒊𝒏 𝟐𝒙) 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝟐 𝒍𝒐𝒈(𝒔𝒊𝒏 𝒙) − 𝒍𝒐𝒈(𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 − 𝒍𝒐𝒈 𝟐 + 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 − 𝒍𝒐𝒈 (𝒄𝒐𝒔 𝒙 ) 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 − 𝒍𝒐𝒈 𝟐 − 𝒍𝒐𝒈 𝒔𝒊𝒏𝒙 − 𝒍𝒐𝒈(𝒄𝒐𝒔 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 − 𝒍𝒐𝒈 𝟐 − 𝒍𝒐𝒈(𝒄𝒐𝒔 𝒙 𝒅𝒙
- 22. න 𝟎 𝝅 𝟐 𝟐 𝒍𝒐𝒈(𝒔𝒊𝒏 𝒙)− 𝒍𝒐𝒈(𝒔𝒊𝒏 𝟐𝒙) 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 − 𝒍𝒐𝒈 𝟐 − 𝒍𝒐𝒈(𝒄𝒐𝒔 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 𝒅𝒙 − න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝟐 𝒅𝒙 − න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒄𝒐𝒔 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 𝒅𝒙 − 𝒍𝒐𝒈 𝟐 න 𝟎 𝝅 𝟐 𝒅𝒙 − න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒄𝒐𝒔 𝝅 𝟐 − 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 𝒅𝒙 − 𝒍𝒐𝒈 𝟐 න 𝟎 𝝅 𝟐 𝒅𝒙 − න 𝟎 𝝅 𝟐 𝒍𝒐𝒈(𝒔𝒊𝒏 𝒙) 𝒅𝒙 = −𝒍𝒐𝒈 𝟐 × 𝒙 𝟎 𝝅 𝟐 = −𝒍𝒐𝒈 𝟐 × 𝝅 𝟐 − 𝟎 න 𝟎 𝝅 𝟐 𝟐 𝒍𝒐𝒈(𝒔𝒊𝒏 𝒙) − 𝒍𝒐𝒈(𝒔𝒊𝒏 𝟐𝒙) 𝒅𝒙 = − 𝝅 𝟐 𝒍𝒐𝒈 𝟐
- 23. 13. Evaluate: 𝟎 𝝅 𝟐 𝒙𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙+𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝝅 𝟐 − 𝒙 𝒔𝒊𝒏 𝝅 𝟐 − 𝒙 𝒄𝒐𝒔 𝝅 𝟐 − 𝒙 𝒔𝒊𝒏 𝟒 𝝅 𝟐 − 𝒙 + 𝒄𝒐𝒔𝟒 𝝅 𝟐 − 𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝝅 𝟐 − 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝟒𝒙 + 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙 න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝝅 𝟐 − 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝟒𝒙 + 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙
- 24. න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = න 𝟎 𝝅 𝟐 𝝅 𝟐 − 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝟒𝒙 + 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙 Each integral = 𝟏 𝟐 (sum of the two integrant) න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 + 𝝅 𝟐 − 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝟒𝒙 + 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙 = 𝟏 𝟐 න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 + 𝝅 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 − 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = 𝝅 𝟒 න 𝟎 𝝅 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙
- 25. න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = 𝝅 𝟒 න 𝟎 𝝅 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 Dividing Numerator and Denominator by 𝒄𝒐𝒔𝟒𝒙 න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = 𝝅 𝟒 න 𝟎 𝝅 𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒄𝒐𝒔𝟒𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = 𝝅 𝟒 න 𝟎 𝝅 𝟐 𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄𝟐𝒙 𝟏 + 𝒕𝒂𝒏𝟒𝒙 𝒅𝒙 Let 𝒖 = 𝒕𝒂𝒏𝟐 𝒙 Let 𝒅𝒖 = 𝟐 𝒕𝒂𝒏𝒙 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 𝐭𝐚𝐧 𝒙 𝒔𝒆𝒄𝟐 𝒙 = 𝟏 𝟐 𝒅𝒖 When 𝒙 = 𝟎, 𝒖 = 𝒕𝒂𝒏𝟐 𝟎 = 𝟎 When 𝒙 = 𝝅 𝟐 , 𝒖 = 𝒕𝒂𝒏𝟐 𝝅 𝟐 = ∞ 𝝅 𝟒 න 𝟎 𝝅 𝟐 𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄𝟐 𝒙 𝟏 + 𝒕𝒂𝒏𝟐𝒙 𝟐 𝒅𝒙 = 𝝅 𝟒 න 𝟎 ∞ 𝟏 𝟐 𝒅𝒖 𝟏 + 𝒖𝟐
- 26. 𝝅 𝟒 න 𝟎 𝝅 𝟐 𝒕𝒂𝒏 𝒙𝒔𝒆𝒄𝟐𝒙 𝟏 + 𝒕𝒂𝒏𝟐𝒙 𝟐 𝒅𝒙 = 𝝅 𝟒 න 𝟎 ∞ 𝟏 𝟐 𝒅𝒖 𝟏 + 𝒖𝟐 = 𝝅 𝟖 𝒕𝒂𝒏−𝟏 (𝒖) 𝟎 ∞ = 𝝅 𝟖 𝒕𝒂𝒏−𝟏 ∞ − 𝒕𝒂𝒏−𝟏 𝟎 = 𝝅 𝟖 𝝅 𝟐 − 𝟎 න 𝟎 𝝅 𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = 𝝅𝟐 𝟏𝟔
- 27. 𝟏𝟒. 𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆: න 𝒙𝟐+ 𝟏 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 − 𝟐 𝒍𝒐𝒈 𝒙 𝒙𝟒 𝒅𝒙 න 𝒙𝟐 + 𝟏 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 − 𝟐 𝒍𝒐𝒈 𝒙 𝒙𝟒 𝒅𝒙 = න 𝒙𝟐 + 𝟏 𝒙𝟒 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 − 𝒍𝒐𝒈 𝒙𝟐 𝒅𝒙 = න 𝒙𝟐 + 𝟏 𝒙𝟒 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 − 𝒍𝒐𝒈 𝒙𝟐 𝒅𝒙 = න 𝒙𝟐 + 𝟏 𝒙 𝟏 𝒙𝟑 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 𝒙𝟐 𝒅𝒙 = න 𝟏 + 𝒙𝟐 𝒙𝟐 𝟏 𝒙𝟑 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 𝒙𝟐 𝒅𝒙
- 28. න 𝒙𝟐 + 𝟏𝒍𝒐𝒈 𝒙𝟐 + 𝟏 − 𝟐 𝒍𝒐𝒈 𝒙 𝒙𝟒 𝒅𝒙 = න 𝟏 + 𝒙𝟐 𝒙𝟐 𝟏 𝒙𝟑 𝒍𝒐𝒈 𝒙𝟐 + 𝟏 𝒙𝟐 𝒅𝒙 = න 𝟏 + 𝟏 𝒙𝟐 𝒍𝒐𝒈 𝟏 + 𝟏 𝒙𝟐 𝟏 𝒙𝟑 𝒅𝒙 Let 𝒕 = 𝟏 + 𝟏 𝒙𝟐 𝒅𝒕 = −𝟐 𝒙𝟑 𝒅𝒙 𝟏 𝒙𝟑 𝒅𝒙 = − 𝟏 𝟐 𝒅𝒕 න 𝟏 + 𝟏 𝒙𝟐 𝒍𝒐𝒈 𝟏 + 𝟏 𝒙𝟐 𝟏 𝒙𝟑 𝒅𝒙 = න 𝒕 𝒍𝒐𝒈 𝒕 −𝟏 𝟐 𝒅𝒕 = − 𝟏 𝟐 න 𝒕 𝒍𝒐𝒈 𝒕 𝒅𝒕
- 29. න 𝟏 + 𝟏 𝒙𝟐 𝒍𝒐𝒈𝟏 + 𝟏 𝒙𝟐 𝟏 𝒙𝟑 𝒅𝒙 = − 𝟏 𝟐 න 𝒕 𝒍𝒐𝒈 𝒕 𝒅𝒕 Let 𝒖 = 𝒍𝒐𝒈 𝒕 𝒅𝒗 = 𝒕 𝒅𝒕 𝒅𝒖 = 𝒅𝒕 𝒕 𝒗 = 𝟐 𝟑 𝒕 𝟑 𝟐 − 𝟏 𝟐 න 𝒕 𝒍𝒐𝒈 𝒕 𝒅𝒕 = − 𝟏 𝟐 𝟐 𝟑 𝒕 𝟑 𝟐 × 𝐥𝐨𝐠 𝒕 − න 𝟐 𝟑 𝒕 𝟑 𝟐 𝒅𝒕 𝒕 = − 𝟏 𝟑 𝒕 𝟑 𝟐 × 𝒍𝒐𝒈 𝒕 + 𝟏 𝟑 න 𝒕 𝟏 𝟐𝒅𝒕 = − 𝟏 𝟑 𝒕 𝟑 𝟐 × 𝒍𝒐𝒈 𝒕 + 𝟏 𝟑 𝟐 𝟑 𝒕 𝟑 𝟐
- 30. − 𝟏 𝟐 න 𝒕 𝒍𝒐𝒈𝒕 𝒅𝒕 = − 𝟏 𝟑 𝒕 𝟑 𝟐 × 𝒍𝒐𝒈 𝒕 + 𝟏 𝟑 𝟐 𝟑 𝒕 𝟑 𝟐 = − 𝟏 𝟑 𝒕 𝟑 𝟐 × 𝒍𝒐𝒈 𝒕 + 𝟐 𝟗 𝒕 𝟑 𝟐 = − 𝟏 𝟑 𝟏 + 𝟏 𝒙𝟐 𝟑 𝟐 × 𝒍𝒐𝒈 𝟏 + 𝟏 𝒙𝟐 + 𝟐 𝟗 𝟏 + 𝟏 𝒙𝟐 𝟑 𝟐 𝒙𝟐+𝟏 𝒍𝒐𝒈 𝒙𝟐+𝟏 −𝟐 𝒍𝒐𝒈 𝒙 𝒙𝟒 𝒅𝒙 = 𝟏 𝟑 𝟏 + 𝟏 𝒙𝟐 𝟑 𝟐 𝟐 𝟑 − 𝒍𝒐𝒈 𝟏 + 𝟏 𝒙𝟐
- 31. 15. A function𝑭(𝒙) is called the anti-derivative or primitive of a given function 𝒇(𝒙) with respect to 𝒙 if 𝒅 𝒅𝒙 𝑭 𝒙 = 𝒇 𝒙 . The anti-derivative of function is not unique. If 𝑭(𝒙) is anti-derivative of 𝒇(𝒙) then 𝑭(𝒙) + 𝑪 is also anti-derivative of 𝒇(𝒙) We represent anti-derivative of 𝒇(𝒙) by 𝒇(𝒙) 𝒅𝒙 a) Find the anti-derivative of 𝒇 𝒙 = 𝒕𝒂𝒏𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 Anti-derivative of 𝒕𝒂𝒏𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 = 𝒕𝒂𝒏𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 𝒅𝒙 = න 𝒔𝒊𝒏𝟑 𝒙 𝒄𝒐𝒔𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 𝒅𝒙 = න 𝒔𝒊𝒏𝟑 𝒙 𝒄𝒐𝒔𝟔𝒙 𝒅𝒙
- 32. Anti-derivative of 𝒕𝒂𝒏𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 = 𝒔𝒊𝒏𝟑𝒙 𝒄𝒐𝒔𝟔𝒙 𝒅𝒙 =න 𝒔𝒊𝒏𝟐 𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔𝟔𝒙 𝒅𝒙 = න 𝟏 − 𝒄𝒐𝒔𝟐𝒙 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔𝟔𝒙 𝒅𝒙 = න 𝟏 𝒄𝒐𝒔𝟔𝒙 − 𝟏 𝒄𝒐𝒔𝟒𝒙 𝒔𝒊𝒏 𝒙 𝒅𝒙 Let 𝒕 = 𝐜𝐨𝐬 𝒙 𝒅𝒕 = −𝒔𝒊𝒏 𝒙 𝒅𝒙 𝒔𝒊𝒏 𝒙 𝒅𝒙 = −𝒅𝒕 න 𝟏 𝒄𝒐𝒔𝟔𝒙 − 𝟏 𝒄𝒐𝒔𝟒𝒙 𝒔𝒊𝒏 𝒙 𝒅𝒙 = න 𝟏 𝒕𝟔 − 𝟏 𝒕𝟒 (−𝒅𝒕)
- 33. න 𝟏 𝒄𝒐𝒔𝟔𝒙 − 𝟏 𝒄𝒐𝒔𝟒𝒙 𝒔𝒊𝒏 𝒙 𝒅𝒙= න 𝟏 𝒕𝟔 − 𝟏 𝒕𝟒 (−𝒅𝒕) = − 𝒕−𝟓 −𝟓 − 𝒕−𝟑 −𝟑 + 𝑪 = 𝟏 𝟓 × 𝟏 𝒕𝟓 − 𝟏 𝟑 × 𝟏 𝒕𝟑 + 𝑪 = 𝟏 𝟓 × 𝟏 𝒄𝒐𝒔𝟓𝒙 − 𝟏 𝟑 × 𝟏 𝒄𝒐𝒔𝟑𝒙 + 𝑪 Anti-derivative of 𝒕𝒂𝒏𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 = 𝒔𝒆𝒄𝟓𝒙 𝟓 − 𝒔𝒆𝒄𝟑𝒙 𝟑 + 𝑪
- 34. b) If theprimitive of 𝒇 𝒙 = 𝒙𝟑 𝟏+𝒙𝟐 is 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪, find the value of 𝒂 and 𝒃 Primitive of 𝒙𝟑 𝟏+𝒙𝟐 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 න 𝒙𝟑 𝟏 + 𝒙𝟐 𝒅𝒙 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 න 𝒙𝟑 𝟏 + 𝒙𝟐 𝒅𝒙 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 න 𝒙𝟐 (𝒙) 𝟏 + 𝒙𝟐 𝒅𝒙 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪
- 35. න 𝒙𝟐(𝒙) 𝟏 + 𝒙𝟐 𝒅𝒙= 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 Let 𝒖 = 𝟏 + 𝒙𝟐 𝒅𝒖 = 𝟐𝒙𝒅𝒙 𝒙 𝒅𝒙 = 𝟏 𝟐 𝒅𝒖 න 𝒖 − 𝟏 𝟏 𝟐 𝒅𝒖 𝒖 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 𝟏 𝟐 න 𝒖 𝒖 − 𝟏 𝒖 𝒅𝒖 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 𝟏 𝟐 න 𝒖 𝟏 𝟐 − 𝒖− 𝟏 𝟐 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪
- 36. 𝟏 𝟐 න 𝒖 𝟏 𝟐 −𝒖− 𝟏 𝟐 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 𝟏 𝟐 𝒖 𝟑 𝟐 𝟑 𝟐 − 𝒖 𝟏 𝟐 𝟏 𝟐 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 𝟏 𝟐 𝟐 𝟑 𝟏 + 𝒙𝟐 𝟑 𝟐 − 𝟐 𝟏 + 𝒙𝟐 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 𝟏 𝟑 𝟏 + 𝒙𝟐 𝟑 𝟐 − 𝟏 + 𝒙𝟐 = 𝒂 𝟏 + 𝒙𝟐 𝟑 𝟐 + 𝒃 𝟏 + 𝒙𝟐 + 𝑪 𝒂 = 𝟏 𝟑 𝒃 = −𝟏