Misc 31 - Definite integral sin 2x tan-1 (sin x) dx - Definate Integration - By Substitution

Slide27.JPG
Slide28.JPG Slide29.JPG Slide30.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
Ask Download

Transcript

Misc 31 Evaluate the definite integral ﷐0﷮﷐𝜋﷮2﷯﷮﷐sin﷮2𝑥﷯﷐﷐tan﷮−1﷯﷮﷐﷐sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 ﷐0﷮﷐𝜋﷮2﷯﷮﷐sin﷮2𝑥﷯﷐﷐tan﷮−1﷯﷮﷐﷐sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷐0﷮﷐𝜋﷮2﷯﷮2﷐sin﷮𝑥﷯﷐cos﷮𝑥﷯ ﷐﷐tan﷮−1﷯﷮﷐﷐sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 Let ﷐sin﷮𝑥﷯=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 ﷐cos﷮𝑥﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥=﷐𝑑𝑡﷮﷐cos﷮𝑥﷯﷯ Substituting x and dx ﷐0﷮﷐𝜋﷮2﷯﷮2﷐sin﷮𝑥﷐cos﷮𝑥 ﷐𝑡𝑎𝑛﷮−1﷯ (﷐sin﷮𝑥) ﷯﷯﷯﷯ 𝑑𝑥 = ﷐0﷮1﷮2𝑡﷐cos﷮𝑥 ﷐𝑡𝑎𝑛﷮−1﷯ (𝑡) ﷯﷯﷐𝑑𝑡﷮𝑐𝑜𝑠𝑥﷯ = ﷐0﷮1﷮2𝑡 ﷐𝑡𝑎𝑛﷮−1﷯𝑡 ﷯ 𝑑𝑡 = 2﷐0﷮1﷮𝑡 ﷐𝑡𝑎𝑛﷮−1﷯ ﷐𝑡﷯﷯ 𝑑𝑡 =2﷐﷐𝑡𝑎𝑛﷮−1﷯𝑡﷐﷮﷮𝑡﷯ 𝑑𝑡 − ﷐﷮﷮﷐﷐𝑑 ﷐﷐𝑡𝑎𝑛﷮−1﷯﷯﷮𝑑𝑡﷯ ﷐﷮﷮𝑡 𝑑𝑡 ﷯﷯ ﷯𝑑𝑡﷯ = 2﷐﷐𝑡𝑎𝑛﷮−1﷯ 𝑡 ﷐﷐﷐𝑡﷮2﷯﷮2﷯﷯−﷐﷮﷮﷐1﷮1 + ﷐𝑡﷮2﷯﷯﷯×﷐﷐𝑡﷮2﷯﷮2﷯ 𝑑𝑡﷯ = 2﷐﷐﷐𝑡﷮2﷯﷮2﷯﷐𝑡𝑎𝑛﷮−1﷯ 𝑡−﷐1﷮2﷯﷐﷮﷮﷐﷐𝑡﷮2﷯﷮2﷯﷯ 𝑑𝑡﷯ = ﷐𝑡﷮2﷯ ﷐𝑡𝑎𝑛﷮−1﷯ 𝑡−﷐﷮﷮﷐﷐𝑡﷮2﷯﷮1 + ﷐𝑡﷮2﷯﷯﷯ 𝑑𝑡 Solving ﷐𝑰﷮𝟏﷯ ﷐I﷮1﷯ = ﷐﷮﷮﷐﷐𝑡﷮2﷯﷮1 + ﷐𝑡﷮2﷯﷯﷯𝑑𝑡 Adding and Subtracting 1 in numerator. ﷐I﷮1﷯ = ﷐﷮﷮﷐﷐﷐𝑡﷮2﷯ + 1 − 1﷮﷐𝑡﷮2﷯ + 1﷯﷯𝑑𝑡﷯ ﷐I﷮1﷯= ﷐﷮﷮﷐﷐﷐﷐𝑡﷮2﷯ + 1﷮﷐𝑡﷮2﷯ + 1﷯−﷐1﷮﷐𝑡﷮2﷯ + 1﷯﷯﷮𝑑𝑡﷯﷯ ﷐I﷮1﷯= ﷐﷮﷮﷐1−﷐1﷮﷐𝑡﷮2﷯ + 1﷯﷯﷯ ﷐I﷮1﷯= ﷐﷮﷮𝑑𝑡−﷐﷮﷮﷐𝑑𝑡﷮﷐𝑡﷮2﷯ + 1﷯﷯﷯ ﷐I﷮1﷯= t − ﷐𝑡𝑎𝑛﷮−1﷯ (t) Thus, our equation becomes ∴ ﷐﷮﷮﷐𝑡𝑎𝑛﷮−1﷯ ﷐𝑡﷯×𝑡 𝑑𝑡﷯= ﷐𝑡﷮2﷯ ﷐𝑡𝑎𝑛﷮−1﷯−( ﷐I﷮1﷯) = ﷐𝑡﷮2﷯﷐𝑡𝑎𝑛﷮−1﷯ ﷐𝑡﷯−﷐𝑡−﷐𝑡𝑎𝑛﷮−1﷯﷐𝑡﷯﷯ = ﷐𝑡﷮2﷯﷐𝑡𝑎𝑛﷮−1﷯ ﷐𝑡﷯−𝑡+﷐𝑡𝑎𝑛﷮−1﷯﷐𝑡﷯ =𝐹(𝑥) 2﷐0﷮1﷮﷐𝑡𝑎𝑛﷮−1﷯ ﷐𝑡﷯ 𝑡 𝑑𝑡﷯ =𝐹﷐1﷯−𝐹(0) =﷐1×﷐𝑡𝑎𝑛﷮−1﷯ ﷐1﷯−1+﷐𝑡𝑎𝑛﷮−1﷯﷐1﷯﷯﷐−0+﷐𝑡𝑎𝑛﷮−1﷯﷐0﷯﷯ = ﷐𝜋﷮4﷯−1+﷐𝜋﷮4﷯−﷐0− 0+0﷯ = ﷐𝜋﷮2﷯−1−0 = ﷐𝝅﷮𝟐﷯−𝟏

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail