Example 27 - Chapter 7 Class 12 Integrals

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Example 27 Evaluate the following integrals: (i) 2 3 2 Step 1 :- 2 = 2 + 1 2 + 1 = 3 3 Hence F = 3 3 Step 2 :- 2 3 2 = 3 2 = 3 3 3 2 3 3 = 27 3 8 3 = 19 3 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) = Example 27 Evaluate the following integrals: (iii) 1 2 + 1 + 2 Step 1 :- = + 1 + 2 We can write the integrate as : + 1 + 2 = A + 1 + B + 2 + 1 + 2 = A + 2 + B + 1 + 1 + 2 By canceling denominators =A +2 +B +1 Therefore + 1 + 2 = 1 + 1 + 2 + 2 Integrating w.r.t. +1 +2 = 1 +1 + 2 +2 = +1 +2 +2 = +1 + +2 2 = +2 2 +1 = + 2 2 + 1 Hence = + 2 2 + 1 Step 2 :- 1 2 + 1 + 2 = 2 1 1 2 + 1 + 2 = 2 + 2 2 2 + 1 1 + 2 2 1 + 1 = 4 2 3 3 2 2 = 4 2 3 3 2 2 = 4 2 3 2 3 2 = 16 3 2 9 = 32 27 = Example 27 Evaluate the following integrals: (iv) 0 4 sin 3 2 cos 2 0 4 3 2 2 Step 1 :- F = 3 2 2 Let s 2 = Differentiating w.r.t. ( sin 2 ) = 2c 2 = = 2 2 Hence the integrate 3 2 2 = 3 2 2 2 = 1 2 3 = 1 2 3+1 3+1 = 1 2 4 4 = 4 8 Putting back = 2 = 1 8 4 2 Hence F = 1 8 4 2 Step 2 :- 0 4 3 2 2 = 4 0 = 1 8 4 2 4 1 8 4 2 0 = 1 8 4 2 1 8 4 0 = 1 8 1 4 1 8 0 4 = 1 8 1 0 = 1 8