Example 27 - Class 12

Transcript

Example 27 Evaluate the following integrals: (i) ﷐2﷮3﷮﷐𝑥﷮2﷯﷯ 𝑑𝑥 Step 1 :- ﷐﷮﷮﷐𝑥﷮2﷯ 𝑑𝑥﷯=﷐﷐𝑥﷮2﷯ + 1 ﷮2 + 1﷯=﷐﷐𝑥﷮3﷯ ﷮3﷯ Hence F﷐𝑥﷯=﷐﷐𝑥﷮3﷯﷮3﷯ Step 2 :- ﷐2﷮3﷮﷐𝑥﷮2﷯ 𝑑𝑥﷯=𝐹﷐3﷯−𝐹﷐2﷯ =﷐﷐3﷮3﷯﷮3﷯−﷐﷐2﷮3﷯﷮3﷯=﷐27﷮3﷯−﷐8﷮3﷯=﷐19﷮3﷯ Example 27 Evaluate the following integrals: (ii) ﷐4﷮9﷮﷐﷐﷮𝑥﷯﷮﷐﷐30 − ﷐𝑥﷮﷐3﷮2﷯﷯﷯﷮2﷯ ﷯﷯ 𝑑𝑥 Step 1 :- ﷐﷮﷮﷐﷐﷮𝑥﷯﷮﷐﷐30 − ﷐𝑥﷮﷐3﷮2﷯﷯﷯﷮2﷯﷯﷯ 𝑑𝑥 Let 30−﷐𝑥﷮﷐3﷮2﷯﷯=𝑡 Differentiating w.r.t. 𝑥 both sides ﷐𝑑﷐30 − ﷐𝑥﷮﷐3﷮2﷯﷯﷯﷮𝑑𝑥﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ −﷐3﷮2﷯﷐𝑥﷮﷐3﷮2﷯ −1﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ −﷐3﷮2﷯﷐𝑥﷮﷐1﷮2﷯ ﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥=﷐𝑑𝑡﷮− ﷐3﷮2﷯﷐ 𝑥﷮ ﷐1﷮2﷯ ﷯﷯ 𝑑𝑥=﷐−2𝑑𝑡﷮3﷐﷮𝑥﷯﷯ Therefore, our equation becomes ﷐﷮﷮﷐﷐﷮𝑥﷯ 𝑑𝑥 ﷮﷐﷐30−﷐𝑥﷮ ﷐3﷮2﷯﷯﷯﷮2﷯﷯=﷐﷮﷮﷐﷐﷮𝑥﷯﷮﷐𝑡﷮2﷯﷯ ﷐−2 𝑑𝑡﷮3 ﷐﷮𝑥﷯﷯﷯﷯ =﷐−2﷮ 3﷯﷐﷮﷮﷐ 𝑑𝑡﷮﷐𝑡﷮2﷯﷯﷯ =﷐−2﷮ 3﷯﷐﷮﷮﷐𝑡﷮2﷯ 𝑑𝑡﷯ =﷐−2﷮ 3﷯ ﷐﷐𝑡﷮− 2 + 1﷯﷮− 2 + 1﷯ =﷐−2﷮ 3﷯ ﷐﷐𝑡﷮− 1﷯﷮−1﷯ =﷐2﷮3﷯ ﷐𝑡﷮−1﷯ =﷐2﷮3𝑡﷯ Putting 𝑡=﷐30−﷐𝑥﷮﷐3﷮2﷯﷯﷯ =﷐2﷮3﷐30 − ﷐𝑥﷮﷐3﷮2﷯﷯﷯﷯ Hence F﷐𝑥﷯=﷐2﷮3﷐30 − ﷐𝑥﷮﷐3﷮2﷯﷯﷯﷯ Step 2 :- ﷐4﷮9﷮﷐﷐﷮𝑥﷯﷮﷐30 − ﷐𝑥﷮ ﷐3﷮2﷯﷯﷯﷯﷯ 𝑑𝑥=𝐹﷐9﷯−𝐹﷐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)﷯ =﷐𝟏𝟗﷮𝟗𝟗﷯ Example 27 Evaluate the following integrals: (iii) ﷐1﷮2﷮﷐𝑥 𝑑𝑥﷮﷐𝑥 + 1﷯ ﷐𝑥 + 2﷯﷯﷯ 𝑑𝑥 Step 1 :- 𝐹﷐𝑥﷯=﷐﷮﷮﷐𝑥 𝑑𝑥﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯﷯ We can write the integrate as : ﷐𝑥﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯=﷐A﷮𝑥 + 1﷯+﷐B﷮𝑥 + 2﷯ ﷐𝑥﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯=﷐A﷐𝑥 + 2﷯ + B﷐𝑥 + 1﷯﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯ By canceling denominators 𝑥=A﷐𝑥+2﷯+B﷐𝑥+1﷯ Therefore ﷐𝑥﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯=﷐−1﷮𝑥 + 1﷯+﷐2﷮𝑥 + 2﷯ Integrating w.r.t.𝑥 ﷐﷮﷮﷐𝑥﷮﷐𝑥+1﷯﷐𝑥+2﷯﷯=﷐﷮﷮﷐−1﷮﷐𝑥+1﷯﷯𝑑𝑥﷯+﷐﷮﷮﷐2﷮𝑥+2﷯𝑑𝑥﷯﷯ =−𝑙𝑜𝑔﷐𝑥+1﷯+2𝑙𝑜𝑔﷐𝑥+2﷯ =−𝑙𝑜𝑔﷐𝑥+1﷯+𝑙𝑜𝑔﷐﷐𝑥+2﷯﷮2﷯ =𝑙𝑜𝑔﷐﷐𝑥+2﷯﷮2﷯−𝑙𝑜𝑔﷐𝑥+1﷯ =𝑙𝑜𝑔﷐﷐﷐﷐𝑥 + 2﷯﷮2﷯﷮𝑥 + 1﷯﷯ Hence 𝐹﷐𝑥﷯=𝑙𝑜𝑔﷐﷐﷐﷐𝑥 + 2﷯﷮2﷯﷮𝑥 + 1﷯﷯ Step 2 :- ﷐1﷮2﷮﷐𝑥﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯𝑑𝑥﷯=𝐹﷐2﷯−𝐹﷐1﷯ ﷐1﷮2﷮﷐𝑥﷮﷐𝑥 + 1﷯﷐𝑥 + 2﷯﷯𝑑𝑥﷯=𝑙𝑜𝑔﷐﷐﷐﷐2 + 2﷯﷮2﷯﷮2 + 1﷯﷯−𝑙𝑜𝑔﷐﷐﷐﷐1 + 2﷯﷮2﷯﷮1 + 1﷯﷯ =𝑙𝑜𝑔﷐﷐﷐﷐4﷯﷮2﷯﷮3﷯﷯−𝑙𝑜𝑔﷐﷐﷐﷐3﷯﷮2﷯﷮2﷯﷯ =𝑙𝑜𝑔﷐﷐﷐﷐4﷮2﷯﷮3﷯﷮﷐﷐3﷮2﷯﷮2﷯﷯﷯ =𝑙𝑜𝑔﷐﷐﷐4﷮2﷯﷮3﷯ × ﷐2﷮﷐3﷮2﷯﷯﷯ =𝑙𝑜𝑔﷐﷐16﷮3﷯ × ﷐2﷮9﷯﷯ =𝑙𝑜𝑔﷐﷐32﷮27﷯ ﷯ =﷐𝐥𝐨𝐠﷮﷐﷐𝟑𝟐﷮𝟐𝟕﷯﷯﷯ Example 27 Evaluate the following integrals: (iv) ﷐0﷮﷐𝜋﷮4﷯﷮﷐﷐sin﷮3﷯﷮2𝑡﷯﷐cos﷮2﷯𝑡﷯ 𝑑𝑡 ﷐0﷮﷐𝜋﷮4﷯﷮𝑠𝑖﷐𝑛﷮3﷯2𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡﷯ Step 1 :- F﷐𝑥﷯=﷐﷮﷮𝑠𝑖﷐𝑛﷮3﷯2𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡﷯ Let s𝑖𝑛 2𝑡=𝑢 Differentiating w.r.t.𝑥 ﷐𝑑(﷐sin﷮2𝑡﷯)﷮𝑑𝑡﷯=﷐𝑑𝑢﷮𝑑𝑡﷯ 2c𝑜𝑠 2𝑡 =﷐𝑑𝑢﷮𝑑𝑡﷯ 𝑑𝑡=﷐𝑑𝑢﷮2 𝑐𝑜𝑠 2𝑡﷯ Hence the integrate ﷐﷮﷮𝑠𝑖﷐𝑛﷮3﷯ 2𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡﷯=﷐﷮﷮﷐𝑢﷮3﷯ 𝑐𝑜𝑠 2𝑡 × ﷐𝑑𝑢﷮2 𝑐𝑜𝑠 2𝑡﷯﷯ =﷐1﷮2﷯﷐﷮﷮﷐𝑢﷮3﷯ 𝑑𝑢﷯ =﷐1﷮2﷯ ﷐﷐𝑢﷮3+1﷯﷮3+1﷯=﷐1﷮2﷯ ﷐﷐𝑢﷮4﷯﷮4﷯= ﷐﷐𝑢﷮4﷯﷮8﷯ Putting back 𝑢=𝑠𝑖𝑛 2𝑡 =﷐1﷮8﷯𝑠𝑖﷐𝑛﷮4﷯ 2𝑡 Hence F﷐𝑡﷯=﷐1﷮8﷯𝑠𝑖﷐𝑛﷮4﷯ 2𝑡 Step 2 :- ﷐0﷮﷐𝜋﷮4﷯﷮𝑠𝑖﷐𝑛﷮3﷯ 2𝑡 𝑐𝑜𝑠 2𝑡=𝐹﷐﷐𝜋﷮4﷯﷯−𝐹﷐0﷯﷯ =﷐1﷮8﷯𝑠𝑖﷐𝑛﷮4﷯ 2﷐﷐𝜋﷮4﷯﷯−﷐1﷮8﷯𝑠𝑖﷐𝑛﷮4﷯ 2﷐0﷯ =﷐1﷮8﷯𝑠𝑖﷐𝑛﷮4﷯﷐𝜋﷮2﷯−﷐1﷮8﷯𝑠𝑖﷐𝑛﷮4﷯ ﷐0﷯ =﷐1﷮8﷯ ×﷐1﷮4﷯−﷐1﷮8﷯ ×﷐0﷮4﷯ =﷐1﷮8﷯ ×1−0 =﷐1﷮8﷯