The value of the integral ∫−π4π4 x+π42−cos⁡2xdx is :

I=∫−π/4π/4 x+π42−cos⁡2xdxI=∫−π/4π/4 −x+π42−cos⁡2xdx ∵∫−aa f(x)dx=∫−aa f(−x)dx⇒2I=∫−π/4π/4 2×π42−cos⁡2xdx⇒I=∫−π/4π/4 π4dx2−cos⁡2x=2×π4∫0π/4 12−cos⁡2xdx ∵∫−aa f(x)dx=2∫0a f(x)dx if f(x) even ⇒I=π2∫0π/4 12−cos⁡2xdx⇒I=π2∫0π/4 12−1−tan2⁡x1+tan2⁡xdxput tan x = t⇒I=π2∫01 dt3t2+1=π23tan−1⁡(3t)01=π23×π3∴I=π263

The value of the integral ∫−π4π4 x+π42−cos⁡2xdx is :

I=∫−π/4π/4 x+π42−cos⁡2xdxI=∫−π/4π/4 −x+π42−cos⁡2xdx ∵∫−aa f(x)dx=∫−aa f(−x)dx⇒2I=∫−π/4π/4 2×π42−cos⁡2xdx⇒I=∫−π/4π/4 π4dx2−cos⁡2x=2×π4∫0π/4 12−cos⁡2xdx ∵∫−aa f(x)dx=2∫0a f(x)dx if f(x) even ⇒I=π2∫0π/4 12−cos⁡2xdx⇒I=π2∫0π/4 12−1−tan2⁡x1+tan2⁡xdxput tan x = t⇒I=π2∫01 dt3t2+1=π23tan−1⁡(3t)01=π23×π3∴I=π263

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The value of the integral $\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x + \frac{π}{4}}{2 - \cos 2 x} dx$ is :

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$\frac{π^{2}}{6 \sqrt{3}}$

$\frac{π^{2}}{12 \sqrt{3}}$

$\frac{π^{2}}{6}$

$\frac{π^{2}}{3 \sqrt{3}}$

answer is A.

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Detailed Solution

$I = \int_{- π / 4}^{π / 4} \frac{x + \frac{π}{4}}{2 - \cos 2 x} dx$
$\begin{array}{l} I = \int_{- π / 4}^{π / 4} \frac{- x + \frac{π}{4}}{2 - \cos 2 x} dx (∵ \int_{- a}^{a} f (x) dx = \int_{- a}^{a} f (- x) dx) \\ \Rightarrow 2 I = \int_{- π / 4}^{π / 4} \frac{2 \times \frac{π}{4}}{2 - \cos 2 x} dx \\ \Rightarrow I = \int_{- π / 4}^{π / 4} \frac{\frac{π}{4} dx}{2 - \cos 2 x} = 2 \times \frac{π}{4} \int_{0}^{π / 4} \frac{1}{2 - \cos 2 x} dx (∵ \int_{- a}^{a} f (x) dx = 2 \int_{0}^{a} f (x) dx if f (x) even) \\ \Rightarrow I = \frac{π}{2} \int_{0}^{π / 4} \frac{1}{2 - \cos 2 x} dx \\ \Rightarrow I = \frac{π}{2} \int_{0}^{π / 4} \frac{1}{2 - \frac{(1 - \tan^{2} x)}{(1 + \tan^{2} x)}} dx \end{array}$
put tan x = t
$\begin{array}{l} {\Rightarrow I = \frac{π}{2} \int_{0}^{1} \frac{dt}{3 t^{2} + 1} = \frac{π}{2 \sqrt{3}} \tan^{- 1} (\sqrt{3} t)]}_{0}^{1} = \frac{π}{2 \sqrt{3}} \times \frac{π}{3} \\ ∴ I = \frac{π^{2}}{6 \sqrt{3}} \end{array}$