Chapter 7.3, Problem 55E

(a)

To determine

To determine the value of ${\displaystyle \int {\cos }^{-1}(x)dx}$

Answer to Problem 55E

The value of the integral ${\displaystyle \int {\cos }^{-1}(x)dx}$ is $={\cos }^{-1}(x).x-\sqrt{1-x^2}+C$

Explanation of Solution

Given information: ${\displaystyle \int {\cos }^{-1}(x)dx}$

Formula used: ${\displaystyle \int f^{-1}(x)dx}={\displaystyle \int y{f}^{\prime }(y)dy}$

Calculation :

Put ${\cos }^{-1}(x)=y$ or $x=\cos (y)$ so that $dx=-\sin (y)dy$

The given integral becomes as:

  ${\displaystyle \int {\cos }^{-1}(x)dx}=-{\displaystyle \int y\sin (y)dy}=y\cos (y)-\sin (y)+C$

  $={\cos }^{-1}(x).x-\sqrt{1-x^2}+C$

(b)

To determine

To determine the value of ${\displaystyle \int {\cos }^{-1}(x)dx}$

Explanation of Solution

Given information: ${\displaystyle \int {\cos }^{-1}(x)dx}$

Formula used: ${\displaystyle \int f^{-1}(x)dx=x{f}^{-1}(x)-{\displaystyle \int x\frac{d}{dx}\left(f^{-1}(x)\right)dx}}$

Calculation :

Using integration by parts, the given integral becomes as:

  ${\displaystyle \int {\cos }^{-1}(x).1dx}={\cos }^{-1}(x).x+{\displaystyle \int \frac{x}{\sqrt{1-x^2}}dx}+C$

Put $1-x^2=t$ so that $-2xdx=dt$

The above expression becomes as:

  $={\cos }^{-1}(x).x-\frac{1}{2}{\displaystyle \int \frac{dt}{\sqrt{t}}}+C$

  $={\cos }^{-1}(x).x-\sqrt{t}+C$

  $={\cos }^{-1}(x).x-\sqrt{1-x^2}+C$

which is the required expression

(c)

To determine

To show that the expression in part (a) and part(b) are same for $C=0$

Explanation of Solution

Given information: ${\displaystyle \int {\cos }^{-1}(x)dx}={\cos }^{-1}(x).x-\sqrt{1-x^2}+C$

Calculation :

For $C=0$ , both the expressions evaluated in part (a) and part (b) are same.

Hence, ${\displaystyle \int {\cos }^{-1}(x)dx}={\cos }^{-1}(x).x-\sqrt{1-x^2}$