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

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

Explanation

Solution

In order to determine the answer of above definite integral use the formula of integration by parts i.e. ∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} } and assume f(x)=cos⁡−1xf(x) = {\cos ^{ - 1}}xandg′(x)=1g'(x) = 1 and calculate f′(x)f'(x)and g(x)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 functioncos⁡−1x dx{\cos ^{ - 1}}x\,dx , whose integral will be
I=∫cos⁡−1xdxI = \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
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} }
In our question Let assume
f(x)=cos⁡−1xf(x) = {\cos ^{ - 1}}x
And g′(x)=1g'(x) = 1
As we know that the derivative of function x is equal to 1
So g(x)=xg(x) = x
and now calculating the derivative of f(x)f(x)with respect to x using rule of
derivativeddx(cos⁡−1x)=−11−x2\dfrac{d}{{dx}}({\cos ^{ - 1}}x) = - \dfrac{1}{{\sqrt {1 - {x^2}} }}
therefore f′(x)=−11−x2f'(x) = - \dfrac{1}{{\sqrt {1 - {x^2}} }}
now putting the values of f(x),f′(x),g(x) and g′(x)f(x),f'(x),g(x)\,and\,g'(x)into the formula
I=∫cos⁡−1x(1)dx=xcos⁡−1x−∫−11−x2(x)dx I=∫cos⁡−1x(1)dx=xcos⁡−1x−∫−x1−x2dx           −(2)  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. ∫−x1−x2dx\int { - \dfrac{x}{{\sqrt {1 - {x^2}} }}dx} use integration by substitution method by substituting u=1−x2u = 1 - {x^2}.Since the derivative of
uuis
du=−2xdx dx=du−2x  du = - 2xdx \\\ dx = \dfrac{{du}}{{ - 2x}} \\\
∫−x1−x2dx ∫−x2xudx \-∫12udu \-∫12u−12du  \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∫xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}
\-u+C \-1−x2+C  \- \sqrt u + C \\\ \- \sqrt {1 - {x^2}} + C \\\
Therefore, the integral of ∫−x1−x2dx\int { - \dfrac{x}{{\sqrt {1 - {x^2}} }}dx} =−1−x2+C - \sqrt {1 - {x^2}} + Cand
putting this value in the equation (2)
I=∫cos⁡−1x(1)dx=xcos⁡−1x−1−x2+CI = \int {{{\cos }^{ - 1}}x(1)dx = } x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + C
Therefore, the integral∫cos⁡−1xdx\int {{{\cos }^{ - 1}}xdx} is equal to xcos⁡−1x−1−x2+Cx{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + Cwhere C is the constant of integration.
Formula:
∫xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}
cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\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)f(x) be a function .Then the family of all its primitives (or antiderivatives)
is called the indefinite integral of f(x)f(x) and is denoted by ∫f(x)dx\int {f(x)} dx
3.The symbol ∫f(x)dx\int {f(x)dx} is read as the indefinite integral of f(x)f(x)with respect to x.
4.Don’t forget to place the constant of integration CC.