- #1
Saq_Lfc
- 2
- 0
∫√(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 (sin(arcsin(3/4χ))xcos(arcsin(3/4χ))) +8/3(arcsin(3/4χ))+k
2χ√(1-16/9x²) +8/3arcsin(3/4χ) +k
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 (sin(arcsin(3/4χ))xcos(arcsin(3/4χ))) +8/3(arcsin(3/4χ))+k
2χ√(1-16/9x²) +8/3arcsin(3/4χ) +k