Question

The value of $\int\limits_{ - 2}^2 {\cos e{c^{ - 1}}\left( {\cos ecx} \right)dx} $
a.1
b.0
c.4
d.2

Answer 1

Hint: We know that $\cos e{c^{ - 1}}\left( {\cos ecx} \right)$= x. And so our integral becomes $\int\limits_{ - 2}^2 {xdx} $ and this can be solved by using the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $.

Complete step-by-step answer:
Here our integrand is $\cos e{c^{ - 1}}\left( {\cos ecx} \right)$
And we know that $\cos e{c^{ - 1}}\left( {\cos ecx} \right)$= x
Hence our integral becomes $\int\limits_{ - 2}^2 {xdx} $
We know that $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
Applying this formula we get,
$ \Rightarrow \int\limits_{ - 2}^2 {xdx} = \left[ {\dfrac{{{x^2}}}{2}} \right]_{ - 2}^2$
When the limits are given we get the value by subtracting the value of lower limit from upper limit
$
   \Rightarrow \left[ {\dfrac{{{2^2}}}{2} - \dfrac{{{{\left( { - 2} \right)}^2}}}{2}} \right] \\
   \Rightarrow \left[ {\dfrac{4}{2} - \dfrac{4}{2}} \right] = 0 \\
$
Hence the required value is 0
The correct option is b.

Note: Like $\cos e{c^{ - 1}}\left( {\cos ecx} \right)$ = x we have
${\sin ^{ - 1}}\left( {\sin x} \right)$= x
${\cos ^{ - 1}}\left( {\cos x} \right) = x$
$se{c^{ - 1}}\left( {secx} \right) = x$
The tangent inverse function ${\tan ^{ - 1}}x$ is an important integral function, but it has no direct method to find it. We shall find the integration of tangent inverse by using the integration by parts method