Example 27 (ii) - Chapter 7 Class 12 Integrals

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Example 27 Evaluate the following integrals: (ii) 4 9 30 3 2 2 Step 1 :- 30 3 2 2 Let 30 3 2 = Differentiating w.r.t. both sides 30 3 2 = 3 2 3 2 1 = 3 2 1 2 = = 3 2 1 2 = 2 3 Therefore, our equation becomes 30 3 2 2 = 2 2 3 = 2 3 2 = 2 3 2 = 2 3 2 + 1 2 + 1 = 2 3 1 1 = 2 3 1 = 2 3 Putting = 30 3 2 = 2 3 30 3 2 Hence F = 2 3 30 3 2 Step 2 :- 4 9 30 3 2 = 9 4 = 2 3 30 9 3 2 2 3 30 4 3 2 = 2 3 30 3 2 2 3 2 3 30 2 2 2 3 = 2 3 1 30 3 3 1 30 2 3 = 2 3 1 30 27 1 30 8 = 2 3 1 3 1 22 = 2 3 22 3 3 22 = 2 3 19 66 = 19 3 (33) =