Example 29 - Chapter 7 Class 12 Integrals (Term 2)

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Example 29 (Method 1) Evaluate ﷐0﷮1﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯ 𝑑𝑥 Step 1 : Let F﷐𝑥﷯=﷐﷮﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1+﷐ 𝑥﷮2﷯﷯﷯𝑑𝑥 Put ﷐﷐tan﷮−1﷯﷮𝑥﷯=𝑡 Differentiating w.r.t.𝑥 ﷐𝑑﷮𝑑𝑥﷯﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ ﷐1﷮1 + ﷐𝑥﷮2﷯﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ Therefore, ﷐﷮﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1+﷐ 𝑥﷮2﷯﷯﷯𝑑𝑥=﷐﷮﷮﷐𝑡﷮1+﷐𝑥﷮2﷯﷯ × ﷐1+﷐𝑥﷮2﷯﷯𝑑𝑡﷯ =﷐﷮﷮𝑡 𝑑𝑡﷯ =﷐﷐𝑡﷮2﷯﷮2﷯ Putting 𝑡=﷐﷐𝑡𝑎𝑛﷮−1﷯﷮𝑥﷯ =﷐﷐﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯﷮2﷯﷮2﷯ Hence 𝐹﷐𝑥﷯=﷐﷐﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯﷮2﷯﷮2﷯ Step 2 : ﷐﷮﷮﷐﷐﷐𝑡𝑎𝑛﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯=𝐹﷐1﷯−F﷐0﷯ =﷐1﷮2﷯﷐﷐﷐﷐tan﷮−1﷯﷮1﷯﷯﷮2﷯ −﷐1﷮2﷯﷐﷐﷐﷐tan﷮−1﷯﷮0﷯﷯﷮2﷯ =﷐1﷮2﷯﷐﷐﷐𝜋﷮4﷯﷯﷮2﷯−﷐1﷮2﷯﷐﷐0﷯﷮2﷯ =﷐1﷮2﷯ ﷐﷐𝜋﷮2﷯﷮16﷯ = ﷐﷐𝝅﷮𝟐﷯﷮𝟑𝟐﷯ Example 29 (Method 2) Evaluate ﷐0﷮1﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯ 𝑑𝑥 Put 𝑡=﷐﷐tan﷮−1﷯﷮𝑥﷯ Differentiating w.r.t.𝑥 ﷐𝑑𝑡﷮𝑑𝑥﷯=﷐𝑑﷮𝑑𝑥﷯﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯ ﷐𝑑𝑡﷮𝑑𝑥﷯=﷐1﷮1 + ﷐𝑥﷮2﷯﷯ ﷐1+﷐𝑥﷮2﷯﷯𝑑𝑡=𝑑𝑥 Hence when value of x varies from 0 to 1, value of t varies from 0 to ﷐𝜋﷮4﷯ Therefore, ﷐0﷮1﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯=﷐0﷮﷐𝜋﷮4﷯﷮﷐𝑡﷮1 + ﷐𝑥﷮2﷯﷯﷯𝑑𝑥 ﷐1+﷐𝑥﷮2﷯﷯𝑑𝑡 =﷐0﷮﷐𝜋﷮4﷯﷮ 𝑡 𝑑𝑡﷯ =﷐﷐﷐﷐𝑡﷮2﷯﷮2﷯﷯﷮0﷮﷐𝜋﷮4﷯﷯ =﷐1﷮2﷯﷐﷐﷐﷐𝜋﷮4﷯﷯﷮2﷯−﷐﷐0﷯﷮2﷯﷯ =﷐1﷮2﷯ × ﷐﷐𝜋﷮2﷯﷮16﷯ = ﷐﷐𝝅﷮𝟐﷯﷮𝟑𝟐﷯