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Question: Evaluate the integral: \(\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sec x}{\sec x+\cos ecx}dx}\). (a) ...

Evaluate the integral:
∫0π2sec⁡xsec⁡x+cos⁡ecxdx\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sec x}{\sec x+\cos ecx}dx}.
(a) π3\dfrac{\pi }{3}
(b) π2\dfrac{\pi }{2}
(c) π4\dfrac{\pi }{4}
(d) π8\dfrac{\pi }{8}

Explanation

Solution

Hint: Assume the integral be equal to ‘II’. Change sec⁡θ\sec \theta and cos⁡ecθ\cos ec\theta into their respective reciprocals. Then use the property of definite integral given by: ∫abf(x)dx=∫abf(a+b−x)dx\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, ∫abf(x)dx=∫abf(a+b−x)dx\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 ‘II’. Therefore,
I=∫0π2sec⁡xsec⁡x+cos⁡ecxdxI=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sec x}{\sec x+\cos ecx}dx}
Now, we use the transformations, sec⁡θ=1cos⁡θ and cosecθ=1sin⁡θ\sec \theta =\dfrac{1}{\cos \theta }\text{ and cosec}\theta =\dfrac{1}{\sin \theta }.
∴I=∫0π21cos⁡x1cos⁡x+1sin⁡xdx =∫0π2cos⁡xsin⁡xcos⁡x(cos⁡x+sin⁡x)dx =∫0π2sin⁡xcos⁡x+sin⁡xdx......................(i) \begin{aligned} & \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{aligned}
Now, using the property, ∫abf(x)dx=∫abf(a+b−x)dx\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx, we get, I=∫0π2sin⁡(π2+0−x)cos⁡(π2+0−x)+sin⁡(π2+0−x)dx =∫0π2sin⁡(π2−x)cos⁡(π2−x)+sin⁡(π2−x)dx \begin{aligned} & 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{aligned}
Using complementary angle rule, sin⁡(π2−θ)=cos⁡θ and cos(π2−θ)=sin⁡θ\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta \text{ and cos}\left( \dfrac{\pi }{2}-\theta \right)=\sin \theta , we have,
I=∫0π2cos⁡xsin⁡x+cos⁡xdx.....................(ii)I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x+\cos x}dx}.....................(ii)
Adding equations (i) and (ii), we get,

& 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{aligned}$$ 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.