Recent content by Saq_Lfc

can u solve it for me I want to check plus how did that 4 outta the integral turned 3

∫√(16-9χ²) dx this is what I tried let 9/16 χ² = sin² then x=4/3(sinu) and u=arcsin(3/4x) and dx=4/3(cosu) du ∫√(16(1-sin²u) )x 4/3(cosu) du ∫16/3 cos²u du ∫8/3(cos2u + 1) du 4/3(sin2u) +8/3 u +k 4/3(2sinucosu) +8/3 u +k 8/3...