Why don’t we bother to state an integration formula that has to do with ${\cos }^{-1}\left(x\right)?$(Hint: Think about the derivatives of ${\cos }^{-1}\left(x\right)$and ${\sin }^{-1}\left(x\right)$What would the integration formula be? Why is it “redundant,” given the integration formula that has to do with${\sin }^{-1}\left(x\right)?$

Given trigonometric functions${\cos }^{-1}\left(x\right)and{\sin }^{-1}\left(x\right)$

Integration is anti derivation of a function

$$\begin{gathered} f\text{'}\left(x\right)=a\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right) \\ =a\left(1\right)+0 \\ =a \end{gathered}$$

Therefore,the anti derivative of a is given by

$\int adx=ax+C$

Here,a is any constant andC is integrating constant

On other hand,

$$\begin{gathered} \int adx=a\int dx \\ =ax+C \end{gathered}$$

Now,derivative of ${\sin }^{-1}\left(x\right)$is given by

$\frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$

Integration of $\frac{1}{\sqrt{1-x^2}}$is given by

$\int \frac{1}{\sqrt{1-x^2}}={\sin }^{-1}x+c$

Again,we know that $$\begin{gathered} {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2} \\ {\cos }^{-1}x=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x \\ \frac{d}{dx}{\cos }^{-1}x=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x\right) \\ \frac{d}{dx}{\cos }^{-1}x=0-\frac{d}{dx}\left({\sin }^{-1}x\right) \\ \frac{d}{dx}{\cos }^{-1}x=-\frac{d}{dx}\left({\sin }^{-1}x\right) \end{gathered}$$

So,derivative of ${\cos }^{-1}x$is equal to the negtive of derivative of$si{n}^{-1}x$

As integration is the anti derivative,

So,integration of$co{s}^{-1}x$ will be equal to the negtive of integration of$si{n}^{-1}x$

That is integration of role="math" localid="1649787622566" $\left(-\int \frac{1}{\sqrt{1-x^2}}\right)$is given by

$\left(-\int \frac{1}{\sqrt{1-x^2}}\right)=-{\cos }^{-1}x+c$