Question

$\underset{\frac{-\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1}{1+\cos 2x}dx$ is equal to

(a) 1

(b) 2

(c) 3

(d) 4

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{\frac{-\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1}{1+\cos 2x}dx \\ \mathrm{Let}f\left(x\right)=\frac{1}{1+\cos 2x} \\ f\left(-x\right)=\frac{1}{1+\cos \left(-2x\right)} \\ =\frac{1}{1+\cos 2x} \\ =f\left(x\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}0,\mathrm{if}f\left(-x\right)=-f\left(x\right)\\ \underset{0}{\overset{a}{2\int }}f\left(x\right)dx,\mathrm{if}f\left(-x\right)=f\left(x\right)\end{array}\right. \\ \mathrm{Thus}, \\ I=2\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1}{1+\cos 2x}dx \\ =2\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1}{2{\cos }^{2}x}dx\left(\mathrm{using}\mathrm{the}\mathrm{identity}:\cos 2x=2{\cos }^{2}x-1\right) \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1}{{\cos }^{2}x}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}{\sec }^{2}xdx \\ ={\left(\tan x\right)}_0^{\frac{\mathrm{\pi }}{4}} \\ =\left(\tan \frac{\mathrm{\pi }}{4}-\tan 0\right) \\ =1-0 \\ =1 \end{gathered}$$

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Hence, the correct option is (a).