Evaluate the definite integral: int_{pi / 4}^ | vedclass

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(D) Let $I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}} \quad \dots (i)$Using the property $\int_a^b

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$\frac{3}{8} \pi$

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(D) Let $I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}ight)}} \quad \dots (i)$Using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$, where $a = \frac{\pi}{4}$ and $b = \frac{\pi}{2}$, we have $a+b-x = \frac{3\pi}{4} - x$.$I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(\frac{3 \pi}{4} - x - \frac{3 \pi}{8}ight)}}$$I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(\frac{3 \pi}{8} - xight)}}$Since $\sin(	heta) = -\sin(-	heta)$, we have $\sin(\frac{3\pi}{8} - x) = -\sin(x - \frac{3\pi}{8})$.$I = \int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{-\sqrt{8} \sin \left(x-\frac{3 \pi}{8}ight)}} = \int_{\pi / 4}^{\pi / 2} \frac{3 e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}ight)} \, dx}{e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}ight)} + 1} \quad \dots (ii)$Adding $(i)$ and $(ii)$:$2I = \int_{\pi / 4}^{\pi / 2} \frac{3(1 + e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}ight)})}{1 + e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}ight)}} \, dx$$2I = \int_{\pi / 4}^{\pi / 2} 3 \, dx = 3[x]_{\pi / 4}^{\pi / 2} = 3(\frac{\pi}{2} - \frac{\pi}{4}) = 3(\frac{\pi}{4}) = \frac{3\pi}{4}$$I = \frac{3\pi}{8}$

Evaluate the definite integral: $\int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}}$

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