$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi }{4}} \frac{dx}{1+\cos 2x}=\\ a) \;\; 1 \; \; b)\;\; 2\; \; c)\; \; 3\; \; d)\; \; 4$

 

Let $\displaystyle f(x)=\frac{1}{1+\cos 2x}\\$ 

By the properties of definite integral we have

$\mathbf{\int_{-a}^{a} f(x) dx} =\left\{\begin{matrix} 2\int_{0}^{a} f(x)dx \; \; if\; \textbf{f is even}, f(-x)=f(x)\\ 0 \;\; \; \; \; \; \; \; \;\; \; \; \; \; \; \: if \: \textbf{ f is odd},f(-x)=-f(x) \end{matrix}\right.$ 

Here $\displaystyle f(-x)=f(x)$ 

$\displaystyle f(-x)=\frac{1}{1+\cos (-2x)}=\frac{1}{1+\cos 2x}=f(x) \; (\because \cos (-x)=\cos x)$ 

Hence 

$\displaystyle f(x)=\frac{1}{1+\cos 2x}$   is an even function

\[\therefore \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx=2\int_{0}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx\]

And

\[\begin{align*} \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx &=2\int_{0}^{\frac{\pi }{4}} \frac{1}{1+2\cos^{2} x-1} \; dx \; \; (\because \cos 2x=2 \cos ^{2} x-1) \\ &= 2\int_{0}^{\frac{\pi }{4}} \frac{1}{2\cos^{2} x} \; dx\; \\ &= \int_{0}^{\frac{\pi }{4}} \frac{1}{\cos^{2} x} \; dx \; (by\; \; cancelling \; 2's)\\ &=\int_{0}^{\frac{\pi }{4}}\sec^{2} x\; dx \\ &= \left [ \tan x \right ]_{0}^{\frac{\pi }{4}}\\ &=\tan \frac{\pi }{4}-\tan 0 \\ &=1 \end{align*}\]

Hence $\displaystyle \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx =1$ 

And option a) 1 is the correct answer

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