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Evaluate the integral ∫ 0/1 sin −1 ( 1+x 2 / 2x )dx using substitution.

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Final answer:

To evaluate the integral ∫ 0/1 sin−1 (1+x²/2x)dx using substitution, substitute u = 1+x²/2x and then differentiate to find du/dx and dx. Substituting these values into the integral allows us to evaluate it using the substitution method.

Step-by-step explanation:

To evaluate the integral ∫ [0,1] sin-1 (1+x2/2x)dx using substitution, let's consider the substitution u = 1+ x2/2x. Differentiating u with respect to x, we get du/dx = (d/dx)(1+x2/2x) = (2x - x2)/(2x2) = 1/x - x/(2x2). Rearranging this equation gives 2x2du = (2x - x2)dx. So, dx = 2x2du / (2x - x2). Substituting these values into the integral, we get: ∫ 0/1 sin-1 (u) / ((2x - x2)du). Now, we can evaluate this integral using the substitution method.

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