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

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

Vedclass · Accepted Answer

$\frac{\pi}{2} - 1$

Vedclass · Answer

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

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

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