The value of the integral int_{0}^{pi / 4} fr | vedclass

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(D) Let $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$, we can rewrite

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$\frac{1}{4} \log _{e} 3$

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(D) Let $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$, we can rewrite the denominator as $3 + 1 - (\sin x - \cos x)^2 = 4 - (\sin x - \cos x)^2$. Thus, $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{4 - (\sin x - \cos x)^2} dx$. Let $t = \sin x - \cos x$, then $dt = (\cos x + \sin x) dx$. When $x = 0$, $t = \sin 0 - \cos 0 = -1$. When $x = \pi / 4$, $t = \sin(\pi / 4) - \cos(\pi / 4) = 0$. Substituting these into the integral, we get $I = \int_{-1}^{0} \frac{dt}{4 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log |\frac{a+x}{a-x}|$, we have: $I = \frac{1}{2(2)} [\log |\frac{2+t}{2-t}|]_{-1}^{0} = \frac{1}{4} [\log |\frac{2+0}{2-0}| - \log |\frac{2-1}{2+1}|]$. $I = \frac{1}{4} [\log 1 - \log (1/3)] = \frac{1}{4} [0 - (-\log 3)] = \frac{1}{4} \log 3$.

The value of the integral $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

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