int_{-pi / 15}^{pi / 15} frac{cos 5 x}{1+e^{5 | vedclass

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Question

(B) Let $I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5 x}{1+e^{5 x}} d x$. Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) +

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{10}$

Vedclass · Answer

(B) Let $I = \int_{-\frac{\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5 x}{1+e^{5 x}} d x$. Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$, we have: $I = \int_{0}^{\frac{\pi}{15}} \left( \frac{\cos 5x}{1+e^{5x}} + \frac{\cos(-5x)}{1+e^{-5x}} \right) dx$. Since $\cos(-5x) = \cos(5x)$ and $\frac{1}{1+e^{-5x}} = \frac{e^{5x}}{e^{5x}+1}$, the expression becomes: $I = \int_{0}^{\frac{\pi}{15}} \left( \frac{\cos 5x}{1+e^{5x}} + \frac{e^{5x} \cos 5x}{1+e^{5x}} \right) dx$. $I = \int_{0}^{\frac{\pi}{15}} \frac{\cos 5x (1+e^{5x})}{1+e^{5x}} dx = \int_{0}^{\frac{\pi}{15}} \cos 5x dx$. $I = \left[ \frac{\sin 5x}{5} \right]_{0}^{\frac{\pi}{15}} = \frac{1}{5} \left( \sin \frac{\pi}{3} - \sin 0 \right)$. $I = \frac{1}{5} \left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{10}$.

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

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