Integration of $\int^\pi_0 \frac{dx}{1+\sin x}$ $$\int^\pi_0 \frac{dx}{1+\sin x}$$Multiplying numerator and denominator by $\sec x$ $$\int^\pi_0 \frac{\sec x \ dx}{\sec x+\tan x}$$Next, I multiplied numerator and denominator by $\sec x+\tan x$ which gives $$\int^\pi_0 \frac{(\sec^2x+\sec x\tan x)\ dx}{(\sec x+\tan x)^2}$$
Making the substitution $t=\sec x+\tan x$ and $dt=(\sec^2x+\sec x\tan x) \ dx$ it becomes $$\int^{-1}_1 \frac{dt}{t^2}$$ Which is $$-\int^{1}_{-1} \frac{dt}{t^2}$$
Which simplifies to give the answer as $2$. But the answer given is $1$. Where am I going wrong?
 A: You need to be a bit careful in multiplying the numerator and denominator by $\sec x$ or by $\sec x+\tan x$, because these expressions are not defined in the middle of the range of integration, at $\frac12\!\pi$. However, you can write the integral as $$\int_0^\pi\frac{\mathrm d x}{1+\sin x}=\lim_{\varepsilon\to0}\left(\int_0^{\frac12\pi-\varepsilon}\frac{\mathrm d x}{1+\sin x}+\int_{\frac12\pi+\varepsilon}^\pi\frac{\mathrm d x}{1+\sin x}\right),$$and continue from there. Your integral $\int_{-1}^1\mathrm d t/t^2$ is undefined ($1/t^2\to\infty$ as $t\to0$).
A simpler way to do it is by using the symmetry of the function to write the integal as$$\int_0^\pi\frac{\mathrm d x}{1+\sin x}=2\int_0^{\frac12\pi}\frac{\mathrm d x}{1+\sin x}=2\int_0^{\frac12\pi}\frac{\mathrm d x}{1+\cos x}=2\int_0^{\frac14\pi}\frac{\mathrm d u}{\cos^2 u},$$where $x=2u$. This then evaluates to $2[\tan u]_0^{\pi/4}=2.$
A: $0 \le \sin x \le 1$ for $0 \le x \le \pi$
$1 \le 1+\sin x \le 2$
$\frac 12 \le \frac{1}{1+\sin x }\le 1$
$\frac {\pi}2 \le \int_0^{\pi} \frac{1}{1+\sin x } dx \le \pi$
1 can't be right.
