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Answer: $\frac{1}{4}\log 3$

Hint: We use indefinite formula then put limits to solve this integral.

Given: $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$

Solution: $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x+1-1} d x$

\begin{aligned} &=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+1-\left(\sin ^{2} x+\cos ^{2} x\right)+\sin 2 x} d x \quad\left[\because 1=\sin ^{2} x+\cos ^{2} x\right] \mid \\ &=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)} d x \quad[\because \sin 2 x=2 \sin x \cdot \cos x] \end{aligned}

\begin{aligned} &=\int_{0}^{\pi / 4}\left(\frac{\sin x+\cos x}{4-(\cos x-\sin x)^{2}}\right) d x \quad\left[\because a^{2}+b^{2}-2 a b=(a+b)^{2}\right] \\ &=\int_{0}^{\pi / 4}\left(\frac{\sin x+\cos x}{2)^{2}-(\cos x-\sin x)^{2}}\right) d x \end{aligned}

Put $\cos x - \sin x = t$

• $\Rightarrow (-\sin x - \cos x)dx = dt$
• $\Rightarrow -(\sin x+\cos x)dx = dt$
• $\Rightarrow (\sin x+\cos x)dx = -dt$
• if x =0 then t = 1 & if$x=\frac{\pi}{4}$  then t = 0

Then,

\begin{aligned} &\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x \\ &=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{(2)^{2}-(\cos x-\sin x)^{2}}\right) d x \\ &=\int_{1}^{0} \frac{1}{(2)^{2}-(t)^{2}}(-d t)=-\int_{1}^{0} \frac{1}{(2)^{2}-(t)^{2}} d t \\ &=\int_{0}^{1} \frac{1}{(2)^{2}-(t)^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x\right. \end{aligned}

\begin{aligned} &=\frac{1}{2 \times 2}\left[\log \left|\frac{2+t}{2-t}\right|\right]_{0}^{1}\\ &=\frac{1}{4}\left[\log \left|\frac{2+1}{2-1}\right|-\log \left|\frac{2+0}{2-0}\right|\right]\\ &=\frac{1}{4}\left[\log \frac{3}{1}-\log \frac{2}{2}\right]\\ &\left.=\frac{1}{4} \log 3-\log 1\right]\\ &=\frac{1}{4}[\log 3-0]\\ &=\frac{1}{4} \log 3\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \log 1=0] \end{aligned}

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