Integrals Giving Inverse Trigonometric Functions
∫ dx/√(9-x²)
We know that ,"\\int \\dfrac{dx}{\\sqrt{a^2-x^2}}=sin^{-1}\\dfrac{x}{a}+ C"
"\\Rightarrow \\int \\dfrac{dx}{\\sqrt{9-x^2}}=\\int \\dfrac{dx}{\\sqrt{3^2-x^2}}= sin^{-1}\\dfrac{x}{3} +C"
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