∫−π/2π/2x+π42+cosxdx=

I=∫−π/2π/2x+π42+cosxdx I=∫0π/2x+π42+cos x+-x+π42+cos(-x)dx ∵ ∫-aa f(x) dx=∫0af(x)+f(-x)dx =∫0π/2π22+cos xdx =π2∫0π/212+cos xdx put tanx2=t⇒ dx=2dt1+t2 x=0 ⇒ t=0 x=π2 ⇒ t=1 I=π2∫0112+1-t21+t2×2dt1+t2 =π2∫012dt2+2t2+1-t2 =π ∫01dtt2+3 =π 13tan-1t301=π3 tan-1 13-0 = π3 π6=π263

∫−π/2π/2x+π42+cosxdx=

I=∫−π/2π/2x+π42+cosxdx I=∫0π/2x+π42+cos x+-x+π42+cos(-x)dx ∵ ∫-aa f(x) dx=∫0af(x)+f(-x)dx =∫0π/2π22+cos xdx =π2∫0π/212+cos xdx put tanx2=t⇒ dx=2dt1+t2 x=0 ⇒ t=0 x=π2 ⇒ t=1 I=π2∫0112+1-t21+t2×2dt1+t2 =π2∫012dt2+2t2+1-t2 =π ∫01dtt2+3 =π 13tan-1t301=π3 tan-1 13-0 = π3 π6=π263

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

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$\frac{8 π \sqrt{3}}{5}$

$\frac{2 π \sqrt{3}}{9}$

$\frac{4 π^{2} \sqrt{3}}{9}$

$\frac{π^{2}}{6 \sqrt{3}}$

answer is D.

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Detailed Solution

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