Question

Evaluate the integral $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx} $.
(A) $0$
(B) $2$
(C) $ - 1$
(D) None of these

Answer 1

Hint: In this problem, we have to evaluate the definite integral. First we will integrate the given function with respect to $x$ and then we will put limits in place of $x$. After simplification, we will get required value. Note that $\int {\cos xdx = \sin x} $.

Complete step-by-step answer:
Here we need to evaluate integral$\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx} $. Let us say $I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx} $. We know that $\int {\cos xdx = \sin x} $. So, by using this formula we can write
$I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx = \left[ {\sin x} \right]_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}}} $
Now we are going to put an upper limit and lower limit in place of $x$. So, we can write
$I = \sin \left( {\dfrac{\pi }{2}} \right) - \sin \left( { - \dfrac{\pi }{2}} \right)$
We know that $\sin \left( { - \theta } \right) = - \sin \theta $. Use this information in the above expression so we can write
$
  I = \sin \left( {\dfrac{\pi }{2}} \right) + \sin \left( {\dfrac{\pi }{2}} \right) \\
   \Rightarrow I = 2\sin \left( {\dfrac{\pi }{2}} \right) \\
 $
We know that the value of $\sin \left( {\dfrac{\pi }{2}} \right)$ is equal to $1$. Substitute this value so we can write
$
  I = 2\left( 1 \right) \\
   \Rightarrow I = 2 \\
   \Rightarrow \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx} = 2 \\
 $
Hence, we can say that option B is correct.
So, the correct answer is “Option B”.

Note: We can solve this problem in another way. If $f\left( { - x} \right) = f\left( x \right)$ then $f\left( x \right)$ is an even function. In the given problem, $\cos \left( { - x} \right) = \cos x$ so we can say that $f\left( x \right) = \cos x$ is an even function. We know that if $f\left( x \right)$ is an even function then $\int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } $. Using this information,
we can write $I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx = 2\int\limits_0^{\dfrac{\pi }{2}} {\cos xdx} } $. After evaluating the integral, we will get the same answer. That is, $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\cos xdx = 2} $. If $f\left( { - x} \right) = - f\left( x \right)$ then $f\left( x \right)$ is an odd function. If $f\left( x \right)$ is an odd function then $\int\limits_{ - a}^a {f\left( x \right)dx = 0} $. This is an important result when we are dealing with integration in which limits are given as $ - a$ to $a$.