My work: I usage integration by parts. U = cos^-1 x ...du = -dx/sqrt(1-x^2) ... V = x ...dv = dx ...

You are watching: Cos−1(x) dx

so integral udv = <1/2cos^-1(1/2) - 0(cos^-1) 0 + \\int 0 ^ 1/2 xdx/(sqrt(1-x^2) = 1/2* pi/3 + (1/2)dx/ sqrt(1-x^2) - 0dx/sqrt(1-x^2).

Is the last component after pi/3 ~ above the ideal track? For some reason, the textbook had actually lower bound 1 and upper bound 3/4 top top the intergal t^-1/2 <-1/2dt> instead? Using integration by components as you did, setup $u = \\cos^-1(x)$ and $dv = dx \\implies v = x$, your indefinite integral must be

$$uv - \\int v\\,du = x\\cos^-1x - \\int \\underbracex_v\\underbrace\\left(\\frac-\\,dx\\sqrt1-x^2\\right)_du$$

It looks favor you managed the integration by parts okay (at the very least the set up, even though you describe $\\int u\\,dv \\;?$. However, it appears you never integrated the integral top top the right before evaluating.

To integrate, you can do this conveniently by substitution $t = 1 - x^2 \\iff dt = -2x \\iff \\frac 12 dt = -x\\,dx)$.

The bounds on this right-side integral readjust under this substitution: when $\\;x = 1/2,\\; t = 1-(1/2)^2 = \\frac 34.\\;$ as soon as $\\;x = 0$, $\\;t = 1$.

\\beginalign\\int_0^1/2 \\cos^-1 x \\, dx & = x \\cos^-1x\\Big|_0^1/2 - \\frac 12 \\int_1^3/4t^-1/2\\,dt\\,\\\\ \\\\ & = x \\cos^-1x\\Big|_0^1/2 - t^1/2\\Big|_1^3/4 \\endalign

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edited Mar 12 \"14 in ~ 13:29
answered Mar 12 \"14 at 13:08 amWhyamWhy
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