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

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(B) To evaluate the integral $I = \int_0^1 {{\cos }^{ - 1}}x\,dx$,we use integration by parts: $\int u\,dv = uv - \int v\,du$.Let $u = {\cos }^{ - 1}x

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(B) To evaluate the integral $I = \int_0^1 {{\cos }^{ - 1}}x\,dx$,we use integration by parts: $\int u\,dv = uv - \int v\,du$.Let $u = {\cos }^{ - 1}x$ and $dv = dx$.Then $du = -\frac{1}{\sqrt{1 - x^2}}dx$ and $v = x$.Applying the formula: $I = [x{\cos }^{ - 1}x]_0^1 - \int_0^1 x \left( -\frac{1}{\sqrt{1 - x^2}} \right) dx$.$I = [x{\cos }^{ - 1}x]_0^1 + \int_0^1 \frac{x}{\sqrt{1 - x^2}} dx$.For the second integral,let $t = 1 - x^2$,so $dt = -2x\,dx$ or $x\,dx = -\frac{1}{2}dt$.When $x=0, t=1$; when $x=1, t=0$.$I = [x{\cos }^{ - 1}x]_0^1 - \frac{1}{2} \int_1^0 \frac{1}{\sqrt{t}} dt = [x{\cos }^{ - 1}x]_0^1 - \frac{1}{2} [2\sqrt{t}]_1^0$.$I = (1 \cdot {\cos }^{ - 1}(1) - 0 \cdot {\cos }^{ - 1}(0)) - (\sqrt{0} - \sqrt{1}) = (0 - 0) - (0 - 1) = 1$.

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

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