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Misc 39 - Prove that 0->1 sin-1 x dx = pi/2 - 1 - Definate Integration - By Formulae

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Misc 39 Prove that ﷐0﷮1﷮﷐﷐sin﷮−1﷯﷮𝑥﷯﷯ 𝑑𝑥=﷐𝜋﷮2﷯−1 Solving L.H.S ﷐0﷮1﷮﷐𝑠𝑖𝑛﷮−1﷯﷐𝑥﷯ 𝑑𝑥﷯ Let I = ﷐﷮﷮﷐𝑠𝑖𝑛﷮−1﷯﷯﷐𝑥﷯ 𝑑𝑥 Let x = sin 𝜃 dx = cos𝜃 d𝜃 Substituting in I I = ﷐﷮﷮﷐𝑠𝑖𝑛﷮−1﷯(﷐sin﷮𝜃) ﷐cos﷮𝜃 𝑑𝜃﷯﷯﷯ = ﷐﷮﷮𝜃﷐cos﷮𝜃 𝑑𝜃﷯﷯ = 𝜃 ﷐﷮﷮﷐cos﷮𝜃 𝑑𝜃−﷐﷮﷮﷐﷐𝑑 ﷐𝜃﷯﷮𝑑𝜃﷯﷐﷮﷮﷐cos﷮𝜃 𝑑𝜃﷯﷯﷯﷯ 𝑑𝜃﷯﷯ = 𝜃 (sin 𝜃) − ﷐﷮﷮﷐1﷯﷐sin﷮𝜃 𝑑𝜃﷯﷯ = 𝜃 sin 𝜃 − ﷐﷮﷮﷐sin﷮𝜃 𝑑𝜃﷯﷯ = 𝜃 sin 𝜃 − (−cos 𝜃) = 𝜃 sin 𝜃 + cos 𝜃 Putting value of 𝜃 Hence I = 𝜃 sin 𝜃 + cos 𝜃 I = ﷐𝑠𝑖𝑛﷮−1﷯﷐𝑥﷯×𝑥+﷐﷮1−﷐𝑥﷮2﷯﷯ = 𝑥 ﷐𝑠𝑖𝑛﷮−1﷯﷐𝑥﷯+ ﷐﷮1−﷐𝑥﷮2﷯﷯ Thus, ﷐﷮﷮﷐𝑠𝑖𝑛﷮−1﷯﷐𝑥﷯ 𝑑𝑥=𝐹﷐𝑥﷯=𝑥﷐𝑠𝑖𝑛﷮−1﷯﷐𝑥﷯+﷐﷮1−﷐𝑥﷮2﷯﷯﷯ Now, ﷐0﷮1﷮﷐𝑠𝑖𝑛﷮−1﷯ ﷐𝑥﷯ 𝑑𝑥=𝐹﷐1﷯−𝐹(0)﷯ = (1) ﷐𝑠𝑖𝑛﷮−1﷯﷐1﷯+﷐﷮1−1﷯−﷐0﷐ 𝑠𝑖𝑛﷮−1﷯﷐0﷯+﷐﷮1−0﷯ ﷯ =﷐𝜋﷮2﷯−(1) =﷐𝝅﷮𝟐﷯−𝟏 = R.H.S Hence, Proved.

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