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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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.