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Example 27 Evaluate the following integrals: (i) 23𝑥2 𝑑𝑥 Step 1 :- 𝑥2 𝑑𝑥=𝑥2 + 1 2 + 1=𝑥3 3 Hence F𝑥=𝑥33 Step 2 :- 23𝑥2 𝑑𝑥=𝐹3−𝐹2 =333−233=273−83=193 Example 27 Evaluate the following integrals: (ii) 49𝑥30 − 𝑥322 𝑑𝑥 Step 1 :- 𝑥30 − 𝑥322 𝑑𝑥 Let 30−𝑥32=𝑡 Differentiating w.r.t. 𝑥 both sides 𝑑30 − 𝑥32𝑑𝑥=𝑑𝑡𝑑𝑥 −32𝑥32 −1=𝑑𝑡𝑑𝑥 −32𝑥12 =𝑑𝑡𝑑𝑥 𝑑𝑥=𝑑𝑡− 32 𝑥 12 𝑑𝑥=−2𝑑𝑡3𝑥 Therefore, our equation becomes 𝑥 𝑑𝑥 30−𝑥 322=𝑥𝑡2 −2 𝑑𝑡3 𝑥 =−2 3 𝑑𝑡𝑡2 =−2 3𝑡2 𝑑𝑡 =−2 3 𝑡− 2 + 1− 2 + 1 =−2 3 𝑡− 1−1 =23 𝑡−1 =23𝑡 Putting 𝑡=30−𝑥32 =2330 − 𝑥32 Hence F𝑥=2330 − 𝑥32 Step 2 :- 49𝑥30 − 𝑥 32 𝑑𝑥=𝐹9−𝐹4 =2330 − 932−2330 − 432 = 23 30 − 3223 −23 30 − 2223 =23130 − 33−130 − 23 =23130 − 27−130 − 8 =2313−122 =2322 − 33 × 22 =23 1966 =193 (33) =𝟏𝟗𝟗𝟗 Example 27 Evaluate the following integrals: (iii) 12𝑥 𝑑𝑥𝑥 + 1 𝑥 + 2 𝑑𝑥 Step 1 :- 𝐹𝑥=𝑥 𝑑𝑥𝑥 + 1𝑥 + 2 We can write the integrate as : 𝑥𝑥 + 1𝑥 + 2=A𝑥 + 1+B𝑥 + 2 𝑥𝑥 + 1𝑥 + 2=A𝑥 + 2 + B𝑥 + 1𝑥 + 1𝑥 + 2 By canceling denominators 𝑥=A𝑥+2+B𝑥+1 Therefore 𝑥𝑥 + 1𝑥 + 2=−1𝑥 + 1+2𝑥 + 2 Integrating w.r.t.𝑥 𝑥𝑥+1𝑥+2=−1𝑥+1𝑑𝑥+2𝑥+2𝑑𝑥 =−𝑙𝑜𝑔𝑥+1+2𝑙𝑜𝑔𝑥+2 =−𝑙𝑜𝑔𝑥+1+𝑙𝑜𝑔𝑥+22 =𝑙𝑜𝑔𝑥+22−𝑙𝑜𝑔𝑥+1 =𝑙𝑜𝑔𝑥 + 22𝑥 + 1 Hence 𝐹𝑥=𝑙𝑜𝑔𝑥 + 22𝑥 + 1 Step 2 :- 12𝑥𝑥 + 1𝑥 + 2𝑑𝑥=𝐹2−𝐹1 12𝑥𝑥 + 1𝑥 + 2𝑑𝑥=𝑙𝑜𝑔2 + 222 + 1−𝑙𝑜𝑔1 + 221 + 1 =𝑙𝑜𝑔423−𝑙𝑜𝑔322 =𝑙𝑜𝑔423322 =𝑙𝑜𝑔423 × 232 =𝑙𝑜𝑔163 × 29 =𝑙𝑜𝑔3227 =𝐥𝐨𝐠𝟑𝟐𝟐𝟕 Example 27 Evaluate the following integrals: (iv) 0𝜋4sin32𝑡cos2𝑡 𝑑𝑡 0𝜋4𝑠𝑖𝑛32𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡 Step 1 :- F𝑥=𝑠𝑖𝑛32𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡 Let s𝑖𝑛 2𝑡=𝑢 Differentiating w.r.t.𝑥 𝑑(sin2𝑡)𝑑𝑡=𝑑𝑢𝑑𝑡 2c𝑜𝑠 2𝑡 =𝑑𝑢𝑑𝑡 𝑑𝑡=𝑑𝑢2 𝑐𝑜𝑠 2𝑡 Hence the integrate 𝑠𝑖𝑛3 2𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡=𝑢3 𝑐𝑜𝑠 2𝑡 × 𝑑𝑢2 𝑐𝑜𝑠 2𝑡 =12𝑢3 𝑑𝑢 =12 𝑢3+13+1=12 𝑢44= 𝑢48 Putting back 𝑢=𝑠𝑖𝑛 2𝑡 =18𝑠𝑖𝑛4 2𝑡 Hence F𝑡=18𝑠𝑖𝑛4 2𝑡 Step 2 :- 0𝜋4𝑠𝑖𝑛3 2𝑡 𝑐𝑜𝑠 2𝑡=𝐹𝜋4−𝐹0 =18𝑠𝑖𝑛4 2𝜋4−18𝑠𝑖𝑛4 20 =18𝑠𝑖𝑛4𝜋2−18𝑠𝑖𝑛4 0 =18 ×14−18 ×04 =18 ×1−0 =18
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