diff. equation.

\hspace{-16}$Find solution of diff. equation $\mathbf{\int_{0}^{\frac{dy}{dx}}\frac{\cos z}{16+9\sin^2 z}dz=\frac{1}{12}\tan^{-1}(x)}$

ans y = x.sin^-1(4x/3)-1/4*√9-16x^2+C

1 Answers

My Solution:

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω