$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $ is equal to

The value of $\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

Evaluate the definite integral: $\int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$

Let $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the correct expression$(s)$ is(are):
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$

Assertion $(A)$: $\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx = 0$,where $[.]$ denotes the greatest integer function.
Reason $(R)$: $2 \sin x$ is a decreasing function in $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

Let $A = \int\limits_0^1 \frac{e^t}{1 + t} \, dt$. Then $\int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$ has the value: