Slide14.JPG

Slide15.JPG
Slide16.JPG
Slide17.JPG

Go Ad-free

Transcript

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.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo