Q117 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

$\int_{-\pi / 15}^{\pi / 15} \frac{\cos 5 x}{1+e^{5 x}} d x=$

  • A
    $\frac{1}{5}$
  • Correct answer
    B
    $\frac{\sqrt{3}}{10}$
  • C
    $\frac{1}{15}$
  • D
    $\frac{1}{10}$

Solution

$\begin{aligned} & I=\int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5 x}{1+e^{5 x}} d x \Rightarrow I=\int_0^{\frac{\pi}{15}} \frac{\cos 5 x}{1+e^{5 x}} d x+\int_0^{\frac{\pi}{15}} \frac{\cos (-5 x)}{1+e^{(5(-x))}} d x \\ = & \int_0^{\frac{\pi}{15}} \frac{\cos 5 x}{1+e^{5 x}} d x+\int_0^{\frac{\pi}{15}} \frac{e^{5 x} \cos (5 x)}{1+e^{5 x}} d x \\ = & \int_0^{\frac{\pi}{15}} \cos 5 x d x=\frac{1}{5}[\sin 5 x]_0^{\frac{\pi}{15}}=\frac{1}{5}\left[\sin \left(\frac{\pi}{3}\right)\right]=\frac{\sqrt{3}}{10} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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