Example 41 - Evaluate integral [root cot x + root tan x] dx - Examples

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  1. Class 12
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Example 41 Evaluate ﷮﷮ ﷮ cot﷮𝑥﷯﷯+ ﷮ tan﷮𝑥﷯﷯﷯﷯𝑑𝑥 ﷮﷮ ﷮ cot﷮𝑥﷯﷯+ ﷮ tan﷮𝑥﷯﷯﷯﷯𝑑𝑥 = ﷮﷮ ﷮ cot﷮𝑥﷯﷯+ 1﷮ ﷮ cot﷮𝑥﷯﷯﷯﷯﷯𝑑𝑥 = ﷮﷮ cot﷮𝑥﷯ + 1﷮ ﷮ cot﷮𝑥﷯﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ ﷮ tan﷮𝑥﷯﷯ cot﷮𝑥﷯+1﷯﷯﷯ 𝑑𝑥 Let tan﷮𝑥﷯= 𝑡﷮2﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. sec﷮2﷯ 𝑥=2𝑡 𝑑𝑡﷮𝑑𝑥﷯ 1+ tan﷮2﷯ 𝑥=2𝑡 . 𝑑𝑡﷮𝑑𝑥﷯ 1+ 𝑡﷮2﷯﷯﷮2﷯=2𝑡 . 𝑑𝑡﷮𝑑𝑥﷯ 1+ 𝑡﷮4﷯=2𝑡 . 𝑑𝑡﷮𝑑𝑥﷯ 1+ 𝑡﷮4﷯﷯ 𝑑𝑥=2𝑡 𝑑𝑡 𝑑𝑥= 2𝑡﷮1 + 𝑡﷮4﷯﷯ . 𝑑𝑡 Putting values of t & dt, we get ﷮﷮ ﷮ 𝑡﷮2﷯﷯ cot﷮𝑥﷯+1﷯﷯﷯ 𝑑𝑥 = ﷮﷮ ﷮ 𝑡﷮2﷯﷯ 1﷮ tan﷮𝑥﷯﷯ +1﷯﷯﷯ 𝑑𝑥 = ﷮﷮𝑡 1﷮ 𝑡﷮2﷯﷯ +1﷯﷯ 𝑑𝑥 = ﷮﷮𝑡 1﷮ 𝑡﷮2﷯﷯ +1﷯﷯ 𝑑𝑥 = ﷮﷮𝑡 1 + 𝑡﷮2﷯﷮ 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮𝑡 1 + 𝑡﷮2﷯﷮ 𝑡﷮2﷯﷯﷯﷯ × 2𝑡﷮1 + 𝑡﷮2﷯﷯ . 𝑑𝑡 = ﷮﷮2 1 + 𝑡﷮2﷯﷮1 + 𝑡﷮4﷯﷯﷯﷯𝑑𝑡 = 2 ﷮﷮ 1 + 𝑡﷮2﷯﷮1 + 𝑡﷮4﷯﷯﷯𝑑𝑡 Dividing numerator and denominator by 𝑡﷮2﷯ = 2 ﷮﷮ 1 + 𝑡﷮2﷯﷮ 𝑡﷮2﷯﷯﷮ 1 + 𝑡﷮4﷯﷮ 𝑡﷮2﷯﷯﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1﷮ 𝑡﷮2﷯﷯ + 1﷮ 1﷮ 𝑡﷮2﷯﷯ + 𝑡﷮2﷯﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1﷮ 𝑡﷮2﷯﷯ + 1﷮ 1﷮ 𝑡﷮2﷯﷯ + 𝑡﷮2﷯﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷮ 𝑡﷮2﷯ + 1﷮ 𝑡﷮2﷯﷯ + 2 − 2﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷮ 𝑡﷯﷮2﷯ + 1﷮𝑡﷯﷯﷮2﷯− 2 𝑡﷯ 1﷮𝑡﷯﷯ + 2﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷮ 𝑡 − 1﷮𝑡﷯﷯﷮2﷯ + 2﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷮ 𝑡 − 1﷮𝑡﷯﷯﷮2﷯ + ﷮2﷯ ﷯﷮2﷯﷯﷯ . 𝑑𝑡 Let 𝑡− 1﷮𝑡﷯=𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 1+ 1﷮ 𝑡﷮2﷯﷯ = 𝑑𝑦﷮𝑑𝑡﷯ 𝑑𝑡 = 𝑑𝑦﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷯﷯ Putting the values of 1﷮t﷯ −t﷯ and dt, we get = 2 ﷮﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷮ 𝑦﷮2﷯ + ﷮2﷯ ﷯﷮2﷯﷯﷯ . 𝑑𝑡 = 2 ﷮﷮ 1 + 1﷮ 𝑡﷮2﷯﷯﷯﷮ 𝑦﷮2﷯ + ﷮2﷯ ﷯﷮2﷯﷯﷯ × 𝑑𝑦﷮ 1 − 1﷮ 𝑡﷮2﷯﷯﷯﷯ = 2 ﷮﷮ 1﷮ 𝑦﷮2﷯ + ﷮2﷯ ﷯﷮2﷯﷯﷯ . 𝑑𝑦 = 2 1﷮ ﷮2﷯﷯ tan﷮−1﷯﷮ 𝑦﷮ ﷮2﷯﷯﷯ +𝐶1﷯ = 2﷮ ﷮2﷯﷯ tan﷮−1﷯﷮ 𝑦﷮ ﷮2﷯﷯﷯ +2𝐶1 = ﷮2﷯ tan﷮−1﷯﷮ 1﷮𝑡﷯ − 𝑡﷮ ﷮2﷯﷯﷯ +2𝐶1 = ﷮2﷯ tan﷮−1﷯﷮ 𝑡﷮2﷯ − 1﷮ ﷮2﷯ 𝑡﷯﷯ +𝐶 = ﷮2﷯ tan﷮−1﷯﷮ tan﷮𝑥﷯ − 1﷮ ﷮2﷯ ﷮ tan﷮𝑥﷯﷯﷯﷯﷯+𝐶 = ﷮𝟐﷯ 𝒕𝒂𝒏﷮−𝟏﷯﷮ 𝒕𝒂𝒏﷮𝒙﷯ − 𝟏﷮ ﷮𝟐 𝒕𝒂𝒏﷮𝒙﷯﷯ ﷯﷯﷯+𝑪

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.