Example 7 (iii) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 7 Find (iii) ∫1▒sin^3𝑥 𝑑𝑥 We know that 𝑠𝑖𝑛 3𝑥=3 𝑠𝑖𝑛𝑥−4 〖𝑠𝑖𝑛〗^3𝑥 4 〖𝑠𝑖𝑛〗^3 𝑥=3 𝑠𝑖𝑛𝑥−𝑠𝑖𝑛3𝑥 〖𝑠𝑖𝑛〗^3 𝑥=(3 𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛3𝑥)/4 ∫1▒sin^3𝑥 𝑑𝑥=∫1▒(3 sin𝑥 − sin3𝑥)/4 𝑑𝑥 =1/4 ∫1▒(3 sin𝑥−sin3𝑥 ) 𝑑𝑥 =1/4 [3∫1▒sin𝑥 𝑑𝑥−∫1▒sin3𝑥 𝑑𝑥] ∫1▒𝒔𝒊𝒏𝟑𝒙 𝒅𝒙 Let 3𝑥=𝑡 3=𝑑𝑡/𝑑𝑥 𝑑𝑥=1/3 𝑑𝑡 ∫1▒𝒔𝒊𝒏𝟑𝒙 𝒅𝒙=∫1▒sin𝑡 . 1/3 𝑑𝑡 =1/3 ∫1▒sin𝑡 . 𝑑𝑡 =1/3 (〖−cos〗𝑡+𝐶1) =−1/3 cos𝑡+1/3 𝐶1 Putting value of 𝑡 =−1/3 cos3𝑥+1/3 𝐶1 ∫1▒𝒔𝒊𝒏𝒙 𝒅𝒙 =−cos𝑥+𝐶2 Thus, ∫1▒sin^3𝑥 𝑑𝑥=1/4 [3∫1▒〖sin𝑥 𝑑𝑥〗−∫1▒sin3𝑥 𝑑𝑥] =1/4 [3(−cos𝑥+𝐶2)−(−1/3 cos3𝑥+1/3 𝐶1)] =1/4 [−3 cos𝑥+3 𝐶2+ 1/3 cos3𝑥+1/3 𝐶1] =1/4 [−3 cos𝑥+1/3 cos3𝑥+(3 𝐶2−1/3 𝐶1)] =(−3)/4 cos𝑥+1/12 cos3𝑥+1/4 (3 𝐶2−1/3 𝐶1) =(−𝟑)/𝟒 𝒄𝒐𝒔𝒙+𝟏/𝟏𝟐 𝒄𝒐𝒔𝟑𝒙+𝑪 ("As" 1/4 (3 𝐶2−1/3 𝐶1)=𝐶)
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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