int_0^1 cos^{-1} x , dx = | vedclass

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(C) Let $I = \int_0^1 (\cos^{-1} x)(1) \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \cos^{-1} x$ and $dv = dx$. Th

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(C) Let $I = \int_0^1 (\cos^{-1} x)(1) \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \cos^{-1} x$ and $dv = dx$. Then $du = -\frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$. $I = [x \cos^{-1} x]_0^1 - \int_0^1 x \left(-\frac{1}{\sqrt{1-x^2}}\right) \, dx$. $I = [x \cos^{-1} x]_0^1 + \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$. For the integral $\int \frac{x}{\sqrt{1-x^2}} \, dx$,let $t = 1-x^2$,then $dt = -2x \, dx$,so $x \, dx = -\frac{1}{2} \, dt$. $\int \frac{x}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2}$. Thus,$I = [x \cos^{-1} x - \sqrt{1-x^2}]_0^1$. Evaluating at the limits: $I = [1 \cdot \cos^{-1}(1) - \sqrt{1-1^2}] - [0 \cdot \cos^{-1}(0) - \sqrt{1-0^2}]$. $I = [1 \cdot 0 - 0] - [0 - 1] = 0 - (-1) = 1$.

$\int_0^1 \cos^{-1} x \, dx =$

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