Ex 7.2, 23 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at May 29, 2018 by Teachoo
Integration by substitution - Trignometric - Inverse
Integration by substitution - Trignometric - Inverse
Last updated at May 29, 2018 by Teachoo
Ex 7.2, 23 sin^(β1)β‘π₯/β(1 β π₯^2 ) Step 1: Let sin^(β1)β‘π₯=π‘ Differentiating both sides π€.π.π‘.π₯ 1/β(1 β π₯^2 )= ππ‘/ππ₯ ππ₯=β(1 β π₯^2 ) .ππ‘ Step 2: Integrating the function β«1βγ" " sin^(β1)β‘π₯/β(1 β π₯^2 ) " " γ. ππ₯ putting sin^(β1)β‘π₯=π‘ & ππ₯=β(1 β π₯^2 ) .ππ‘ = β«1βγπ‘/β(1 β π₯^2 ) " " γ. β(1 β π₯^2 ) .ππ‘ = β«1βγπ‘ . ππ‘γ = π‘^2/2+πΆ = (γπππγ^(βπ)β‘π )^π/π+πͺ