int_{0}^{frac{pi}{4}} frac{sin x+cos x}{9+16 | vedclass

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Question

(B) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x = k \log 3$. Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin

Vedclass · Accepted Answer

$\frac{1}{20}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x = k \log 3$. Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$. When $x = 0$,$t = -1$. When $x = \frac{\pi}{4}$,$t = 0$. Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$. Substituting these into the integral: $I = \int_{-1}^{0} \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^{0} \frac{dt}{25 - 16t^2} = \frac{1}{16} \int_{-1}^{0} \frac{dt}{(\frac{5}{4})^2 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log |\frac{a+x}{a-x}|$: $I = \frac{1}{16} 	imes \frac{1}{2(\frac{5}{4})} [\log |\frac{\frac{5}{4} + t}{\frac{5}{4} - t}|]_{-1}^{0} = \frac{1}{40} [\log |\frac{5+4t}{5-4t}|]_{-1}^{0}$. Evaluating the limits: $I = \frac{1}{40} [\log(1) - \log(\frac{5-4}{5+4})] = \frac{1}{40} [0 - \log(\frac{1}{9})] = \frac{1}{40} \log(9) = \frac{1}{40} \log(3^2) = \frac{2}{40} \log 3 = \frac{1}{20} \log 3$. Comparing with $k \log 3$,we get $k = \frac{1}{20}$.

$\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x=k \log 3$,then $k=$

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