1. We need to evaluate $I=80\int_0^{\pi/4}\frac{\sin\theta+\cos\theta}{9+16\sin2\theta}\,d\theta.$

2. Use the standard substitution $t=\sin\theta-\cos\theta.$ Then, $dt=(\cos\theta+\sin\theta)\,d\theta.$ This is very useful because the numerator is exactly $(\sin\theta+\cos\theta)\,d\theta$.

Also, $(\sin\theta-\cos\theta)^2=\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=1-\sin2\theta.$ Hence, $\sin2\theta=1-t^2.$ So the denominator becomes $9+16\sin2\theta=9+16(1-t^2)=25-16t^2.$

Therefore, $I=80\int \frac{dt}{25-16t^2}$ with limits converted as follows:

• When $\theta=0$, $t=\sin0-\cos0= -1.$
• When $\theta=\pi/4$, $t=\sin\frac\pi4-\cos\frac\pi4=0.$

Thus, $I=80\int_{-1}^{0}\frac{dt}{25-16t^2}.$

3. Now factor the denominator: $25-16t^2=(5-4t)(5+4t).$ Use the standard result $\int \frac{dt}{a^2-b^2t^2}=\frac{1}{2ab}\ln\left(\frac{a+bt}{a-bt}\right)+C.$ Here $a=5$, $b=4$, so $\int \frac{dt}{25-16t^2}=\frac{1}{40}\ln\left(\frac{5+4t}{5-4t}\right)+C.$

Therefore, $I=80\left[\frac{1}{40}\ln\left(\frac{5+4t}{5-4t}\right)\right]_{-1}^{0}.$

4. Simplify: $I=2\left[\ln\left(\frac{5+4t}{5-4t}\right)\right]_{-1}^{0}.$ At $t=0$, $\ln\left(\frac{5}{5}\right)=\ln 1=0.$ At $t=-1$, $\ln\left(\frac{5-4}{5+4}\right)=\ln\left(\frac{1}{9}\right)=-\ln 9.$ So, $I=2\big(0-(-\ln 9)\big)=2\ln 9=4\ln 3.$

5. Hence the value of the integral is $\boxed{4\log 3}$ (where $\log$ denotes natural logarithm).

6. Option check:

• A: $3\log 4$ ❌
• B: $4\log 3$ ✅
• C: $6\log \frac43$ ❌
• D: $2\log 3$ ❌

So the correct option is B.