Evaluate the integral:
$\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sec x}{\sec x+\cos ecx}dx}$.
(a) $\dfrac{\pi }{3}$
(b) $\dfrac{\pi }{2}$
(c) $\dfrac{\pi }{4}$
(d) $\dfrac{\pi }{8}$

Hint: Assume the integral be equal to ‘$I$’. Change $\sec \theta $ and $\cos ec\theta $ into their respective reciprocals. Then use the property of definite integral given by: $\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx$ to simplify the integral.

Complete step-by-step answer:
Here, we have been provided with a definite integral. There are certain properties of definite integral but here we will use a basic property which is, $\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx$.
Now, let us come to the question. Let us assume the given integral is ‘$I$’. Therefore,
\[I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sec x}{\sec x+\cos ecx}dx}\]
Now, we use the transformations, $\sec \theta =\dfrac{1}{\cos \theta }\text{ and cosec}\theta =\dfrac{1}{\sin \theta }$.
$\begin{align}
  & \therefore I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}dx} \\
 & =\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{\cos x\left( \cos x+\sin x \right)}dx} \\
 & =\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\cos x+\sin x}dx}......................(i) \\
\end{align}$
Now, using the property, $\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx$, we get, \[\begin{align}
  & I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( \dfrac{\pi }{2}+0-x \right)}{\cos \left( \dfrac{\pi }{2}+0-x \right)+\sin \left( \dfrac{\pi }{2}+0-x \right)}dx} \\
 & =\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( \dfrac{\pi }{2}-x \right)}{\cos \left( \dfrac{\pi }{2}-x \right)+\sin \left( \dfrac{\pi }{2}-x \right)}dx} \\
\end{align}\]
Using complementary angle rule, $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta \text{ and cos}\left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $, we have,
\[I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x+\cos x}dx}.....................(ii)\]
Adding equations (i) and (ii), we get,
\[\begin{align}
  & 2I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\cos x+\sin x}dx}+\int_{0}^{\dfrac{\pi }{2}}{\dfrac{co\operatorname{s}x}{\sin x+\cos x}dx} \\
 & \text{ }=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\cos x+\sin x}dx} \\
 & \text{ }=\int_{0}^{\dfrac{\pi }{2}}{dx} \\
 & \text{ }=\left[ x \right]_{0}^{\dfrac{\pi }{2}} \\
 & \text{ }=\left( \dfrac{\pi }{2}-0 \right) \\
 & \text{ }=\dfrac{\pi }{2} \\
 & \therefore I=\dfrac{1}{2}\times \dfrac{\pi }{2} \\
 & \text{ }=\dfrac{\pi }{4} \\
\end{align}\]
Hence, option (c) is the correct answer.

Note: Properties of definite integrals are very important. Here, we have used one of the properties of definite integral and it became so easy to simplify. If we will not use properties of definite integral here and solve it like an indefinite integral then it will be a very lengthy and time consuming process. So, basic properties of definite integral are important to solve this question.