Example 29 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at May 29, 2018 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
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Example 3 (ii) Important
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Example 4
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Example 5 (iii) Important
Example 5 (iv) Important
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Example 6 (ii) Important
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Example 7 (ii) Important
Example 7 (iii)
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Example 8 (ii) Important
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Example 9 (ii) Important
Example 9 (iii) Important
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Example 10 (ii) Important
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Example 20 Important
Example 21 Important
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Example 24
Example 25 Important Deleted for CBSE Board 2022 Exams
Example 26 Deleted for CBSE Board 2022 Exams
Example 27 (i)
Example 27 (ii) Important
Example 27 (iii)
Example 27 (iv) Important
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Example 29 You are here
Example 30 Important
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Example 39 Important
Example 40 Important
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Example 44 Important
Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
Chapter 7 Class 12 Integrals (Term 2)
Serial order wise
Transcript
Example 29 (Method 1) Evaluate 01tan−1𝑥1 + 𝑥2 𝑑𝑥 Step 1 : Let F𝑥=tan−1𝑥1+ 𝑥2𝑑𝑥 Put tan−1𝑥=𝑡 Differentiating w.r.t.𝑥 𝑑𝑑𝑥tan−1𝑥=𝑑𝑡𝑑𝑥 11 + 𝑥2=𝑑𝑡𝑑𝑥 Therefore, tan−1𝑥1+ 𝑥2𝑑𝑥=𝑡1+𝑥2 × 1+𝑥2𝑑𝑡 =𝑡 𝑑𝑡 =𝑡22 Putting 𝑡=𝑡𝑎𝑛−1𝑥 =tan−1𝑥22 Hence 𝐹𝑥=tan−1𝑥22 Step 2 : 𝑡𝑎𝑛−1𝑥1 + 𝑥2=𝐹1−F0 =12tan−112 −12tan−102 =12𝜋42−1202 =12 𝜋216 = 𝝅𝟐𝟑𝟐 Example 29 (Method 2) Evaluate 01tan−1𝑥1 + 𝑥2 𝑑𝑥 Put 𝑡=tan−1𝑥 Differentiating w.r.t.𝑥 𝑑𝑡𝑑𝑥=𝑑𝑑𝑥tan−1𝑥 𝑑𝑡𝑑𝑥=11 + 𝑥2 1+𝑥2𝑑𝑡=𝑑𝑥 Hence when value of x varies from 0 to 1, value of t varies from 0 to 𝜋4 Therefore, 01tan−1𝑥1 + 𝑥2=0𝜋4𝑡1 + 𝑥2𝑑𝑥 1+𝑥2𝑑𝑡 =0𝜋4 𝑡 𝑑𝑡 =𝑡220𝜋4 =12𝜋42−02 =12 × 𝜋216 = 𝝅𝟐𝟑𝟐
