Question

How do you find the integral of \[{\cos ^{ - 1}}x\,dx\]?

Answer 1

Hint:In order to determine the answer of above definite integral use the formula of integration by parts i.e. $\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $ and assume $f(x) = {\cos ^{ - 1}}x$and$g'(x) = 1$ and calculate $f'(x)$and $g(x)$and put into the formula and use the substitution method to find the integral of the second term in the formula . Putting back all in the original equation, you will get your required integral.

Complete step by step solution:
We are given a function\[{\cos ^{ - 1}}x\,dx\] , whose integral will be
$I = \int {{{\cos }^{ - 1}}xdx} $
We will use integration by parts method to find the integral of the above.
The formula for calculation of integration of parts is
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $
In our question Let assume
$f(x) = {\cos ^{ - 1}}x$
And $g'(x) = 1$
As we know that the derivative of function x is equal to 1
So $g(x) = x$
and now calculating the derivative of $f(x)$with respect to x using rule of
derivative$\dfrac{d}{{dx}}({\cos ^{ - 1}}x) = - \dfrac{1}{{\sqrt {1 - {x^2}} }}$
therefore $f'(x) = - \dfrac{1}{{\sqrt {1 - {x^2}} }}$
now putting the values of $f(x),f'(x),g(x)\,and\,g'(x)$into the formula
$
I = \int {{{\cos }^{ - 1}}x(1)dx = } x{\cos ^{ - 1}}x - \int { - \dfrac{1}{{\sqrt {1 - {x^2}} }}(x)dx}
\\
I = \int {{{\cos }^{ - 1}}x(1)dx = } x{\cos ^{ - 1}}x - \int { - \dfrac{x}{{\sqrt {1 - {x^2}} }}dx}
\,\,\,\,\,\,\,\,\,\,\, - (2) \\
$
So to calculate the integral of the second term in the formula i.e. $\int { - \dfrac{x}{{\sqrt {1 - {x^2}}
}}dx} $use integration by substitution method by substituting $u = 1 - {x^2}$.Since the derivative of
$u$is
$
du = - 2xdx \\
dx = \dfrac{{du}}{{ - 2x}} \\
$
$
\int { - \dfrac{x}{{\sqrt {1 - {x^2}} }}dx} \\
\int { - \dfrac{x}{{2x\sqrt u }}dx} \\
- \int {\dfrac{1}{{2\sqrt u }}du} \\
- \int {\dfrac{1}{2}{u^{ - \dfrac{1}{2}}}du} \\
$
Using the rule of integral$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $
$
- \sqrt u + C \\
- \sqrt {1 - {x^2}} + C \\
$
Therefore, the integral of $\int { - \dfrac{x}{{\sqrt {1 - {x^2}} }}dx} $=$ - \sqrt {1 - {x^2}} + C$and
putting this value in the equation (2)
$I = \int {{{\cos }^{ - 1}}x(1)dx = } x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + C$
Therefore, the integral$\int {{{\cos }^{ - 1}}xdx} $ is equal to $x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} +
C$where C is the constant of integration.
Formula:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $
\[{\cos ^2}x + {\sin ^2}x = 1\]
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } $
Additional Information:
1.Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction
2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.

Note:
1.Use standard formula carefully while evaluating the integrals.
2. Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives)
is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
3.The symbol $\int {f(x)dx} $ is read as the indefinite integral of $f(x)$with respect to x.
4.Don’t forget to place the constant of integration $C$.