The integral $80\displaystyle\int_0^{\pi/4}\left(\frac{\sin\theta+\cos\theta}{9+16\sin 2\theta}\right)d\theta$ is equal to:
Step-by-Step Solution
Key Concept: Substitute $t = \sin\theta - \cos\theta$ so $(\cos\theta + \sin\theta)d\theta = dt$ and $\sin 2\theta = 1-t^2$; limits shift from $t=-1$ to $t=0$, giving a standard $\int 1/(a^2-t^2)\,dt$ form.
Let $t = \sin\theta - \cos\theta$, so $(\cos\theta+\sin\theta)d\theta = dt$ and $\sin 2\theta = 1-t^2$.
Limits: $\theta=0\to t=-1$, $\theta=\pi/4\to t=0$.
$$I = \int_{-1}^{0}\frac{dt}{9+16(1-t^2)} = \frac{1}{16}\int_{-1}^{0}\frac{dt}{\frac{25}{16}-t^2}.$$
$$= \frac{1}{4}\left[\frac{1}{10}\ln\frac{5+4t}{5-4t}\right]_{-1}^{0} = \frac{1}{40}\ln 9 = \frac{\ln 9}{40}.$$
$$80I = 2\ln 9 = 4\ln 3 = 4\log_e 3.$$
Correct Answer: 2