Mathematics

Evaluate the following : $$\int _{ -\frac { \pi  }{ 4 }  }^{ \frac { \pi  }{ 4 }  }{ \dfrac { x+\dfrac { \pi  }{ 4 }  }{ 2-\cos 2x } dx } $$


SOLUTION
It is take $$\displaystyle I = \int_{-\pi/4}^{\pi/4}\dfrac{x + \pi/4}{2-\cos 2x}dx$$      ...(i)

use property $$\displaystyle \int_a^b f(x) dx = \int_a^b f(a+b - x)dx$$

Then 

$$I =\displaystyle  \int_{-\pi/4}^{\pi/4} \dfrac{-x+\pi/4}{2-\cos 2x}dx$$      ...(2) 

adding (1) and (2) we have 
$$2I =\displaystyle \int_{-\pi/4}^{\pi/4}\dfrac{\pi/2}{2-\cos 2x}dx = \dfrac{\pi}{2} \int_{-\pi/4} ^{\pi/4} \dfrac{dx}{1 + 2\sin^2x}$$

$$I = \displaystyle \dfrac{\pi}{4}\int_{-\pi/4} ^{\pi/4} \dfrac{dx}{1+2\sin^2x} = \dfrac{\pi}{2} \int_0^{\pi/4} \dfrac{dx}{1 + 2\sin^2x}$$

$$I = \dfrac{\pi}{2} \displaystyle \int_{0}^{\pi/4} \dfrac{\sec^2xdx}{\sec^2x + 2\tan^2x} = \dfrac{\pi}{2} \int_0^{\pi/4} \dfrac{\sec^2x dx}{3\tan^2x + 1}$$

Let $$\sqrt{3} \tan x = t \Rightarrow \sec^2x\, dx = \dfrac{dt}{\sqrt{3}}$$

$$\left.\begin{matrix} I = \dfrac{\pi}{2\sqrt{3}} \displaystyle \int_0^{\sqrt{3}} \dfrac{dt}{t^2 + 1} = \dfrac{\pi}{2\sqrt{3}} \tan^{-1} t \end{matrix}\right|^{\sqrt{3}}_0$$

$$= \dfrac{\pi}{2\sqrt{3}} \times \dfrac{\pi}{3} = \dfrac{\pi^2}{6\sqrt{3}}$$
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Subjective Medium Published on 17th 09, 2020
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