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

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Aicharya · Accepted Answer

Let $I=\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x=K \log 3$ Put $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$ ...(1) Now squaring equation (1), we get $\begin{aligned} 	herefore I &=\int_{-1}^{0} \frac{d t}{9+16\left(1-t^{2}ight)} \ &=\int_{-1}^{0} \frac{d t}{25-16 t^{2}}=\frac{1}{16} \int_{-1}^{0} \frac{d t}{\left(\frac{5}{4}ight)^{2}-t^{2}} \end{aligned}$ $\begin{array}{l} =\frac{1}{16} 	imes \frac{1}{2\left(\frac{5}{4}ight)}\left[\log \mid \frac{\frac{5}{4}+t}{\frac{5}{4}-t}ight]_{-1}^{0}=\frac{1}{40}\left[\log \left(\frac{5+4 t}{5-4 t}ight)ight]_{-1}^{0}=\frac{1}{40}\left[\log (1)-\log \left(\frac{1}{9}ight)ight] \ =\frac{1}{40}(\log 9)=\frac{1}{40} \log 3^{2}=\frac{2}{40} \log 3=\frac{1}{20} \log 3 \end{array}$ As per given data, $K=\frac{1}{20}$

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

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