Question

Find the value of integration of the given function: $\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{1 + \cos 2x}} = }$
a) 1.
b) 2.
c) 3.
d) 4.

Answer 1

Hint: In this question, we will use some basic properties of definite integrals and some basic trigonometric identities. $\int\limits_{ - a}^a {f(x)dx} = 2\int\limits_0^a {f(x)dx}$, if f is an even function, i.e. f(-x) = f(x).

Complete step-by-step answer:
We know that,
$\cos 2x = 2{\cos ^2}x - 1$.
This can also be written as:
$1 + \cos 2x = 2{\cos ^2}x$. ………..(i)
According to the property number 7 of definite integrals,
$\int\limits_{ - a}^a {f(x)dx} = 2\int\limits_0^a {f(x)dx}$, if f is an even function, i.e. f(-x) = f(x).
And
$\int\limits_{ - a}^a {f(x)dx} = 0$, if f is an odd function, i.e. f(-x) = -f(x).
Let us check for the given function either it is even function or odd function.
So,
$f(x) = \dfrac{1}{{1 + \cos 2x}}$.
For f(-x),
$f( - x) = \dfrac{1}{{1 + \cos ( - 2x)}}$,
And we know that cos is an even function, i.e. $\cos ( - t) = \cos (t)$.
Therefore,
$f( - x) = \dfrac{1}{{1 + \cos (2x)}} = f(x)$.
Now, we can say that the given function is an even function.
Thus, we will apply the above property.
By applying the property for even function, we get
$\Rightarrow \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{1 + \cos 2x}} = } 2\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{1 + \cos 2x}}}$.
Now, using equation(i), we get
$\Rightarrow 2\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{1 + \cos 2x}}} = 2\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{2{{\cos }^2}x}}}$
We know that, $\sec x = \dfrac{1}{{\cos x}}$
So we can write this also as,
$\Rightarrow 2\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{1 + \cos 2x}}} = \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^2}xdx}$
Now integrating this, we get
$\Rightarrow \left[ {\tan x} \right]_0^{\dfrac{\pi }{4}} \\\ \Rightarrow \left[ {\tan \dfrac{\pi }{4} - \tan 0} \right] \\\ \Rightarrow 1 \\\$
Hence we can say that $\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{{dx}}{{1 + \cos 2x}} = } 1$
Therefore, the correct answer is option (1).

Note: Whenever we ask such types of questions, we have to remember the properties of definite integral and also we have to remember the trigonometric properties. First we have to find out whether we can integrate the given function or not. Then we will simplify that function by using some trigonometric identities so that we can integrate it. After that we will apply the required properties of definite integral. By solving it, we will get the required answer.