Question

The value of $\int {\cos e{c^{ - 1}}x dx, x > 1.}$

Answer 1

Here, Given the function$\int {\cos e{c^{ - 1}}x dx, x > 1.}$ to integrate. To solve this function we have to use the method known as Integration by Parts. The formula for this method is $u\int v dx - \int {\dfrac{{du}}{{dx}}} \int {vdx}$. By putting the values of functions in this formula we will get the solution.

Complete step-by-step solution:
Here, Given the trigonometric equation and we need to solve it.
Given equation is $\int {\cos e{c^{ - 1}}x,dx,x > 1.}$ or we can write it as $\int {1*\cos e{c^{ - 1}}xdx}$.
While using the method of Integration by parts we have to separate the $u$& $v$ values of the equation using the LIATE rule.
Where,
L= Logarithmic
I= Inverse Trigonometric
A= Algebraic
T= Trigonometric
E= Exponential
By which serially the value of $u$&$v$ are elected. That means the value of $u$ is $\cos e{c^{ - 1}}x$ because it is Inverse Trigonometric & value of $v$ is 1 or ${x^0}$ which is Algebraic. (We know${x^0}$=1.)
So,
$u$=$\cos e{c^{ - 1}}x$ , $v$ = 1
Substitute these values in formula,
Equation become,
$\Rightarrow$ $\cos e{c^{ - 1}}x\int {1dx - \int {\dfrac{{d\left( {\cos e{c^{ - 1}}x} \right)}}{{dx}}} } \int {1dx}$ _ _ _ _ _ _ _ _ _ _(1)
Some Important formulas,
$\Rightarrow$ $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
$\Rightarrow$ $\int {{x^0}} dx = \dfrac{{{x^{0 + 1}}}}{{0 + 1}} = x$ and,
$\Rightarrow \dfrac{{d\left( {\cos e{c^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - 1}}{{x\sqrt {{x^2} - 1} }}$
So, by substituting these values in equation (1),
(1)$\Rightarrow$ $x\cos e{c^{ - 1}}x - \int {\dfrac{{ - 1}}{{x\sqrt {{x^2} - 1} }}} \times xdx$
$\Rightarrow x\cos e{c^{ - 1}}x + \int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}}$
We know,
$\int {\dfrac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = \log |x + \sqrt {{x^2} - {a^2}} | + c$
$\Rightarrow x\cos e{c^{ - 1}}x + \log |x + \sqrt {{x^2} - 1} | + c$
Therefore, the solution of $\int {\cos e{c^{ - 1}}x dx, x > 1.}$ is $\Rightarrow x\cos e{c^{ - 1}}x + \log |x + \sqrt {{x^2} - 1} | + c$.

Note: Note that we cannot directly integrate inverse trigonometric functions. We have to use some relations and formulas. Here the most important part is selecting the values of u and v. If we select the wrong values of u and v, we will get the wrong answer.