$$ \begin{aligned} \int \frac{-1}{\sqrt{1-x^{2}}} d x & =\cos ^{-1} x+C \ & ext { Or } \ & -[?]^{-1} x+c\end{aligned} $$

Question

$$ \begin{aligned} \int \frac{-1}{\sqrt{1-x^{2}}} d x & =\cos ^{-1} x+C \ & 	ext { Or } \ & -[?]^{-1} x+c\end{aligned} $$

UpStudy AI · Accepted Answer

We begin with the given integral

$$
\int \frac{-1}{\sqrt{1-x^{2}}}\,dx.
$$

Notice that the derivative of the inverse cosine function is

$$
\frac{d}{dx}\bigl(\cos^{-1}x\bigr)=-\frac{1}{\sqrt{1-x^2}},
$$

so one antiderivative is

$$
\cos^{-1}x + C.
$$

Alternatively, consider the inverse sine function whose derivative is

$$
\frac{d}{dx}\bigl(\sin^{-1}x\bigr)=\frac{1}{\sqrt{1-x^2}}.
$$

Thus, the derivative of

$$
-\sin^{-1}x
$$

is

$$
-\frac{d}{dx}\bigl(\sin^{-1}x\bigr)=-\frac{1}{\sqrt{1-x^2}}.
$$

This shows that another antiderivative is

$$
-\sin^{-1}x + C.
$$

So, filling in the missing function, we have

$$
\int \frac{-1}{\sqrt{1-x^{2}}}\,dx=\cos^{-1}x+C \quad \text{or} \quad -\sin^{-1}x+C.
$$
The integral simplifies to $$ \cos^{-1}x + C $$ or $$ -\sin^{-1}x + C $$.

\( \begin{aligned} \int \frac{-1}{\sqrt{1-x^{2}}} d x & =\cos ^{-1} x+C \\ & \text { Or } \\ & -[?]^{-1} x+c\end{aligned} \)

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