If int left{ cos^{-1} x - (1-x^2)^{-frac{1}{2 | vedclass

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(B) Let $I = \int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx$. Given that $\int \left\{ \cos^{-1} x - \frac{1}{\sqrt{1-x^2}} \right

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$e^x$

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(B) Let $I = \int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx$. Given that $\int \left\{ \cos^{-1} x - \frac{1}{\sqrt{1-x^2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$. Let $f(x) = \cos^{-1} x$. Then $f'(x) = -\frac{1}{\sqrt{1-x^2}}$. The integral becomes $\int k \left\{ f(x) + f'(x) \right\} dx = k \cdot \cos^{-1} x + c$. We know that $\int e^x \{ f(x) + f'(x) \} dx = e^x f(x) + c$. Comparing this with the given equation $\int k \{ f(x) + f'(x) \} dx = k \cdot f(x) + c$,we can see that $k$ must be a function of $x$ such that $k = e^x$. Thus,$k = e^x$.

If $\int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$,then $k = $ . . . . . . .

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