Question

How do you find the integral of \[{\cos ^{ - 1}}x\] by parts from $\left[ {0,\dfrac{1}{2}} \right]$?

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. Determine limits by subtracting the value of integral at upper limit and same at lower limit.

Complete step by step solution:
We are given a function so which we have to find the integral using Integration by parts from $\left[ {0,\dfrac{1}{2}} \right]$
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
$
\Rightarrow \int {{{\cos }^{ - 1}}x(1)dx = } x{\cos ^{ - 1}}x - \int { - \dfrac{1}{{\sqrt {1 - {x^2}}
}}(x)dx} \\
\Rightarrow \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)
$ \Rightarrow \int {{{\cos }^{ - 1}}x(1)dx = } x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} $
Assigning limits to the above integral from $\left[ {0,\dfrac{1}{2}} \right]$
Limits are calculated as [Upper limit – lower limit]
$
\Rightarrow \int\limits_0^{\dfrac{1}{2}} {{{\cos }^{ - 1}}xdx = } {\left[ {x{{\cos }^{ - 1}}x - \sqrt {1
- {x^2}} } \right]^{\dfrac{1}{2}}}_0 \\
= \dfrac{1}{2}{\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) - \sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}}
- \left( {0{{\cos }^{ - 1}}(0) - \sqrt 1 } \right) \\
= \dfrac{1}{2}.\dfrac{\pi }{3} - \sqrt {1 - \left( {\dfrac{1}{4}} \right)} + 1 \\
= \dfrac{\pi }{6} - \sqrt {\dfrac{3}{4}} + 1 \\
= \dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{2} + 1 \\
= \dfrac{{\pi + 6 - 3\sqrt 3 }}{6} \\
$
Therefore, the integral $\int\limits_0^{\dfrac{1}{2}} {{{\cos }^{ - 1}}xdx} $ is equal to $\dfrac{{\pi + 6 - 3\sqrt 3 }}{6}$.
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.