Example 6 (i) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Integration by substitution - Trignometric - Normal
Example 5 (i)
Ex 7.2, 22 Important
Misc 15
Example 35
Ex 7.2, 27
Ex 7.2, 31
Ex 7.2, 30
Ex 7.2, 26 Important
Ex 7.2, 24
Example 6 (i) You are here
Ex 7.2, 29 Important
Ex 7.2, 21
Ex 7.2, 34 Important
Ex 7.2, 39 (MCQ) Important
Ex 7.2, 25
Ex 7.2, 32 Important
Ex 7.2, 33 Important
Misc 7 Important
Integration by substitution - Trignometric - Normal
Last updated at Dec. 16, 2024 by Teachoo
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