The value of the integral - 4 4 x + 4 2 - cos 2 x d x is [2023]

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Solution

Q.
$The value of the integral \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - \cos 2 x} d x is$ [2023]

1 $\frac{π^{2}}{12 \sqrt{3}}$

2 $\frac{π^{2}}{6}$

3 $\frac{π^{2}}{3 \sqrt{3}}$

4 $\frac{π^{2}}{6 \sqrt{3}}$

Ans.
(4)

$I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - \cos 2 x} d x ...(i)$

Replace $x$ by $- x$ , we get

$I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{- x + \frac{π}{4}}{2 - \cos 2 x} d x ...(ii)$

Adding (i) and (ii), we get

$2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{π / 2}{2 - \cos 2 x} d x \Rightarrow I = \frac{π}{4} \cdot 2 \int_{0}^{\frac{π}{4}} \frac{d x}{2 - \cos 2 x} d x$

$\Rightarrow I = \frac{π}{4} \cdot 2 \int_{0}^{\frac{π}{4}} \frac{(1 + \tan^{2} x) d x}{2 (1 + \tan^{2} x) - (1 - \tan^{2} x)}$

$= \frac{π}{2} \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x d x}{2 (1 + \tan^{2} x) - (1 - \tan^{2} x)}$

Let $\tan x = t \Rightarrow \sec^{2} x d x = d t$

at $x = 0, t = 0$ and $x = \frac{π}{4}, t = 1$

$I = \frac{π}{2} \int_{0}^{1} \frac{d t}{3 t^{2} + 1} \Rightarrow I = \frac{π}{6} \int_{0}^{1} \frac{d t}{t^{2} + {(\frac{1}{\sqrt{3}})}^{2}} = \frac{π}{6} \sqrt{3} \cdot {[\tan^{- 1} \sqrt{3} t]}_{0}^{1}$

$\Rightarrow I = \frac{π}{2 \sqrt{3}} \tan^{- 1} \sqrt{3} I = \frac{π^{2}}{6 \sqrt{3}}$

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