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Example 6 (i) - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Aug. 20, 2021 by Teachoo
Transcript
Example 6 Find the following integrals: (i) ∫1▒〖sin^3𝑥 cos^2𝑥 〗 𝑑𝑥 ∫1▒〖sin^3𝑥 cos^2𝑥 〗 𝑑𝑥 Let cos 𝑥=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. −sin𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=(−𝑑𝑡)/sin𝑥 Now are equation becomes ∫1▒〖sin^3𝑥 cos^2𝑥 〗 𝑑𝑥 Putting value of 𝑐𝑜𝑠𝑥 and 𝑑𝑥 = ∫1▒sin^3𝑥 .𝑡^2. 𝑑𝑥 = ∫1▒sin^3𝑥 .𝑡^2. 𝑑𝑡/(−sin𝑥 ) = ∫1▒sin^3𝑥/(−sin𝑥 ) 𝑡^2. 𝑑𝑡 = –∫1▒sin^2𝑥 𝑡^2. 𝑑𝑡 = – ∫1▒(1−cos^2𝑥 ) 𝑡^2. 𝑑𝑡 = – ∫1▒(1−𝑡^2 ) 𝑡^2. 𝑑𝑡 = – ∫1▒(𝑡^2−𝑡^4 ) 𝑑𝑡 = ∫1▒(−𝑡^2+𝑡^4 ) 𝑑𝑡 = ∫1▒〖−𝑡^2 〗. 𝑑𝑡 + ∫1▒𝑡^4 . 𝑑𝑡 (∴ sin^2𝑥=1−cos^2𝑥) = (〖−𝑡〗^2+1)/(2 + 1)+𝑡^(4 + 1)/(4 + 1)+𝐶 = (−𝑡^3)/3 +𝑡^5/5 +𝐶 Putting back value of t = cos x = (−𝟏)/𝟑 〖𝒄𝒐𝒔〗^𝟑𝒙 +𝟏/𝟓 〖𝒄𝒐𝒔〗^𝟓𝒙 +𝑪
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Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i) You are here
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Example 6 (iii) Important
Example 7 (i)
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Example 7 (iii)
Example 8 (i)
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Example 9 (i)
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Example 10 (i)
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Example 23
Example 24
Example 25 Important Deleted for CBSE Board 2022 Exams
Example 26 Deleted for CBSE Board 2022 Exams
Example 27 (i)
Example 27 (ii) Important
Example 27 (iii)
Example 27 (iv) Important
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Example 37
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Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.