Example 29 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at May 29, 2018 by
Last updated at May 29, 2018 by
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
Definite Integration - By Substitution
Definite Integration - By Substitution
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