Misc 37 - Prove that 0->1 sin-1 x dx = pi/2 - 1 - Miscellaneous - Miscellaneous

part 2 - Misc 37 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Misc 37 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals part 4 - Misc 37 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals

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Misc 37 Prove that ∫_0^1β–’sin^(βˆ’1)⁑π‘₯ 𝑑π‘₯=πœ‹/2βˆ’1 Solving L.H.S ∫_0^1▒〖〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯) 𝑑π‘₯γ€— Let I = ∫1▒〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯) 𝑑π‘₯ Let x = sin πœƒ dx = cosπœƒ dπœƒ Substituting in I I = ∫1▒〖〖𝑠𝑖𝑛〗^(βˆ’1) (sinβ‘γ€–πœƒ) cosβ‘γ€–πœƒ π‘‘πœƒγ€— γ€— γ€— = ∫1β–’γ€–πœƒ cosβ‘γ€–πœƒ π‘‘πœƒγ€— γ€— = πœƒ ∫1β–’cosβ‘γ€–πœƒ π‘‘πœƒβˆ’βˆ«1β–’((𝑑 (πœƒ))/π‘‘πœƒ ∫1β–’cosβ‘γ€–πœƒ π‘‘πœƒγ€— ) π‘‘πœƒγ€— = πœƒ (sin πœƒ) βˆ’ ∫1β–’γ€–(1) sinβ‘γ€–πœƒ π‘‘πœƒγ€— γ€— = πœƒ sin πœƒ βˆ’ ∫1β–’sinβ‘γ€–πœƒ π‘‘πœƒγ€— = πœƒ sin πœƒ βˆ’ (βˆ’cos πœƒ) = πœƒ sin πœƒ + cos πœƒ Putting value of πœƒ Hence I = πœƒ sin πœƒ + cos πœƒ I = 〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯)Γ—π‘₯+√(1βˆ’π‘₯^2 ) = π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯)+ √(1βˆ’π‘₯^2 ) Thus, ∫1▒〖〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯) 𝑑π‘₯=𝐹(π‘₯)=π‘₯〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯)+√(1βˆ’π‘₯^2 )γ€— Now, ∫1_0^1▒〖〖𝑠𝑖𝑛〗^(βˆ’1) (π‘₯) 𝑑π‘₯=𝐹(1)βˆ’πΉ(0)γ€— = (1) 〖𝑠𝑖𝑛〗^(βˆ’1) (1)+√(1βˆ’1)βˆ’(0γ€– 𝑠𝑖𝑛〗^(βˆ’1) (0)+√(1βˆ’0) ) =πœ‹/2βˆ’(1) =𝝅/πŸβˆ’πŸ = R.H.S Hence, Proved.

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