$$\int^\pi_0 \frac{dx}{1+\sin x}$$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?
Asked
Active
Viewed 137 times
4
-
Just as an alternative approach... Did you consider starting by multiplying by $1-\sin x$ on top and bottom? – abiessu Apr 24 '16 at 05:59
-
You didn't go wrong. The answer is 2. – almagest Apr 24 '16 at 06:05
-
Your work is correct and the book is wrong. By the way, tangent half-angle substitution makes is very fast $\int_0^\infty \frac{2}{(t+1)^2},dt$. – Claude Leibovici Apr 24 '16 at 06:27
-
I am not sure if this kind of notation is considered acceptable. But $1/t^2$ is not integrable at $0$. – velut luna Apr 24 '16 at 06:27
-
Actually, it's $\int_1^\infty + \int_{-\infty}^{-1}$, right? – velut luna Apr 24 '16 at 06:28
2 Answers
5
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 $2[\tan u]_0^{\pi/4}=2.$
John Bentin
- 18,454
4
$0 \le \sin x \le 1$ 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.
tomi
- 9,594