Miscellaneous

Chapter 7 Class 12 Integrals
Serial order wise

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Misc 37 Prove that β«_0^1βsin^(β1)β‘π₯ ππ₯=π/2β1 Solving L.H.S β«_0^1βγγπ ππγ^(β1) (π₯) ππ₯γ Let I = β«1βγπ ππγ^(β1) (π₯) ππ₯ Let x = sin π dx = cosπ dπ Substituting in I I = β«1βγγπ ππγ^(β1) (sinβ‘γπ) cosβ‘γπ ππγ γ γ = β«1βγπ cosβ‘γπ ππγ γ = π β«1βcosβ‘γπ ππββ«1β((π (π))/ππ β«1βcosβ‘γπ ππγ ) ππγ = π (sin π) β β«1βγ(1) sinβ‘γπ ππγ γ = π sin π β β«1βsinβ‘γπ ππγ = π sin π β (βcos π) = π sin π + cos π Putting value of π Hence I = π sin π + cos π I = γπ ππγ^(β1) (π₯)Γπ₯+β(1βπ₯^2 ) = π₯ γπ ππγ^(β1) (π₯)+ β(1βπ₯^2 ) Thus, β«1βγγπ ππγ^(β1) (π₯) ππ₯=πΉ(π₯)=π₯γπ ππγ^(β1) (π₯)+β(1βπ₯^2 )γ Now, β«1_0^1βγγπ ππγ^(β1) (π₯) ππ₯=πΉ(1)βπΉ(0)γ = (1) γπ ππγ^(β1) (1)+β(1β1)β(0γ π ππγ^(β1) (0)+β(1β0) ) =π/2β(1) =π/πβπ = R.H.S Hence, Proved.