Is there another way to solve $\int_{0}^{\pi}\frac{1-\sin x}{\sin x+1}dx$

Question

My way to solve this integral. I wonder is there another way to solve it as it's very long for me.

$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$

Let $$u=\tan (\frac{x}{2})$$ $$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$

By Weierstrass Substitution

$$\sin (x)=\frac{2u}{u^2+1}$$

$$\cos (x)=\frac{1-u^2}{u^2+1}$$

$$dx=\frac{2du}{u^2+1}$$

$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$

$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$

$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$

$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$

$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$

$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$

$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$

$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$

$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$

Let $$s=u+1$$

$$ds=du$$

$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$

$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$

$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$

$$=4-\pi$$

$$\approx 0.85841$$

Answer 1

Substitute $x=\pi/2-2t$ so the integral becomes $$ -2\int_{\pi/4}^{-\pi/4}\frac{1-\cos 2t}{1+\cos 2t}\,dt= 2\int_{-\pi/4}^{\pi/4}\frac{1-\cos^2t}{\cos^2t}\,dt =2\Bigl[\tan t - t\Bigr]_{-\pi/4}^{\pi/4} $$

Answer 2

First use partial fractions to get rid of the sine in the numerator: $$ \int_0^{\pi} \frac{1-\sin{x}-1+1}{1+\sin{x}} \, dx = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, dx = -\pi + \int_0^{\pi} \frac{2 \, dx}{1+\sin{x}}. $$ We have the identity $$ 1+\sin{x} = 2\sin^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} $$ (from $\cos{2\theta}=1-2\sin^2{\theta}$ and $\sin{\theta}=-\cos{(\theta+\pi/2)}$), so the remaining integral is $$ \int_0^{\pi} \csc^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \, dx = \left[ -2 \cot{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \right]_0^{\pi} = -2(-1-1) = 4 $$

Answer 3

Multiply numerator and denominator by $1 - \sin x$. So that $\displaystyle\int_0^\pi \dfrac{1-\sin x}{1+\sin x} \cdot \dfrac{1-\sin x}{1-\sin x}dx $

$$ = \displaystyle\int_0^\pi \dfrac{1-2\sin x+ \sin^2x}{\cos^2x}dx = \int_0^\pi \sec^2x - 2\dfrac{\sin x}{\cos^2x} + \tan^2x dx$$

We know that $\tan^2x + 1 = \sec^2x$. So the integral would look like: $$= \int_0^\pi 2\sec^2x -1 - 2\dfrac{\sin x}{\cos^2x}dx$$

We know the integral of $\sec^2x$ is just $\tan x$, and the last integrand can be solved by letting $u=\cos x$.

Answer 4

Hint: Note that by symmetry our integral is twice the integral from $0$ to $\pi/2$. Then by symmetry replace $\sin x$ by $\cos x$. Then use the identities $\cos x=2\cos^2 (x/2)-1=1-2\sin^2(x/2)$.

Answer 5

\begin{align} A &= \int_{0}^{\pi}\left(\:\frac{1\:-\:\sin (x)}{\sin (x)\:+\:1}\:\right)dx \\ F(x) &= \int\left(\:\frac{1\:-\:\sin (x)}{\sin(x)\:+\:1 }\cdot\frac{\sin(x)\:-\:1}{\sin (x)\:-\:1 }\:\right)dx \\ &= \int\left(\:\frac{ -\sin^2(x) + 2\cdot sin(x)-1}{\sin^2(x)\:-\:1 }\:\right)dx \\ &= \int\left(\:\frac{ -(\sin^2(x) - 2\cdot sin(x)+1)}{-(1-\sin^2(x)) }\:\right)dx \\ &= \int\left(\:\frac{ (\sin^2(x) - 2\cdot sin(x)+1)}{(1-\sin^2(x)) }\:\right)dx \\ &= \int\left(\:\frac{\sin^2(x) - 2\cdot sin(x)+1}{\cos^2(x) }\:\right)dx \\ &= \int\left(\:\frac{\sin^2(x)}{\cos^2(x)}-\frac{ 2\cdot sin(x)}{\cos^2(x)}+\frac{1}{\cos^2(x)}\:\right)dx \\ &= \int\left(\:\frac{\sin^2(x)}{\cos^2(x)}\right)dx-\int\left(\frac{2\cdot sin(x)}{\cos^2(x)}\right)dx+\int\left(\frac{1}{\cos^2(x)}\:\right)dx \\ F(x) &= \tan(x) - x - 2\cdot\sec(x) + \tan(x)\\ A &= F(\pi) - F(0)\\ F(\pi) - F(0) &= (\tan(\pi) - \pi - 2\cdot\sec(\pi) + \tan(\pi))-(\tan(0) - 0 - 2\cdot\sec(0) + \tan(0))\\ &= (0 - \pi - 2\cdot-1 + 0)-(0 - 0 - 2\cdot1 + 0)\\ &= (-\pi + 2)-(-2)\\ &= -\pi + 2+2\\ &= 4-\pi\\ \end{align}

Answer 6

The question asked "is there another way", so here's another one. Start by removing the trig from the numerator:

$$ \int_0^{\pi} \frac{1-\sin{x}}{1+\sin{x}} \, \mathrm{d}x = \int_0^{\pi} \frac{2 - (1 +\sin{x})}{1+\sin{x}} \, \mathrm{d}x = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, \mathrm{d}x = -\pi + \int_0^{\pi} \frac{2 \, \mathrm{d}x}{1+\sin{x}} $$

There are many ways to solve this but I would like to demonstrate the substitution $u = 1 + \sin x$. It would be much more convenient if the limits ran from $0$ to $\frac{\pi}{2}$, so that $x = \sin^{-1} (u-1)$ is a continuous, monotonic function from $[1,2]$ to $[0,\frac{\pi}{2}]$, and also because $\cos x$ and $\sin x$ are both positive for $0 \leq x \leq \frac{\pi}{2}$. The latter property is often useful so we can write $\cos x=\sqrt{1-\sin^2 x}$ and $\sin x=\sqrt{1-\cos^2 x}$. Fortunately symmetry considerations allow a change of limits:

$$\int_0^{\pi} \frac{2 \, \mathrm{d}x}{1+\sin{x}} = \int_0^{\pi/2} \frac{4 \, \mathrm{d}x}{1+\sin{x}}$$

We have $\frac{\mathrm{d}u}{\mathrm{d}x} = \cos x = \sqrt{1-\sin^2 x} = \sqrt{1-(u-1)^2} = \sqrt{2u - u^2}$ so the integral becomes:

$$\int_0^{\pi/2} \frac{4 \, \mathrm{d}x}{1+\sin{x}} = \int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{2u - u^2}} = \int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{-u(u-2)}}$$

This might look fearsome but is actually very amenable to the third Euler substitution, since we have a factorised quadratic expression inside the square root. The computation is very similar to this answer. In general the substitution is $\sqrt{au^2 + bu + c} = \sqrt{a(u-\alpha)(u-\beta)} = (u-\alpha)t$ which gives $u = \frac{a\beta-\alpha t^2}{a-t^2}$; in our case we can take $a=-1$, $b=2$, $c=0$, $\alpha=0$, and $\beta=2$ with $\sqrt{-u(u-2)} = ut$.

Since $u=\frac{(-1)(2)-0t^2}{-1-t^2}=2(t^2+1)^{-1}$ we obtain $\frac{\mathrm{d}u}{\mathrm{d}t} = -4t(t^2+1)^{-2}$ and to change the limits we set $t=\frac{\sqrt{-u(u-2)}}{u} = \sqrt{\frac{2-u}{u}}$. The remaining integral becomes:

$$\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{-u(u-2)}} = \int_1^0 \frac{4 (-4t)(t^2+1)^{-2} \, \mathrm{d}t}{u^2 t} = \int_0^1 \frac{4 (4t)(t^2+1)^{-2} \, \mathrm{d}t}{4(t^2+1)^{-2} t} = \int_0^1 4 \, \mathrm{d}t = 4$$

This was not the most straightforward way to find the result $4 - \pi$, but I just wanted to draw out the similarity between this integral and one the original poster had asked about before (the main differences being the trig in the numerator - which is easily removed - and the limits).

Answer 7

Contour integration is quite a good idea in this case. On the unit circle, we have

$$z=e^{ix}=\cos x + i \sin x$$ and therefore $$\sin x=\frac{z-z^{-1}}{2i}$$ and $$dz=iz\,dx$$ Substitution makes the integral

$$I=\int_C \frac{1}{zi}\frac{-z^2+2iz+1}{z^2+2iz-1}dz=-\int_C\frac{1}{zi}\frac{(z-i)^2}{(z+i)^2}dz=$$ $$=i\int_C\frac{1}{z}\left(\frac{z^2-1-2zi}{z^2+1}\right)^2dz$$ where the integration contour $C$ is the upper half of the unit arc (clockwise). Now you can consider integration around the upper half of the unit disc. The integration splits symbolically as

$$\text{residue}=\int_{-1}^1+\int_C$$ The residue unfortunately lies directly on the integration path, at $z=0$, and is equal to $\operatorname{Res}=i$. As the path is straight at the residue, you can take the Cauchy principal value of the integral and use half the residue:

$$I=\frac12 2\pi i (i)-i\int_{-1}^1 \frac{1}{z}\left(\frac{z^2-1-2zi}{z^2+1}\right)^2dz$$ $$=-\pi-i\int_{-1}^1 \frac{1}{z}\left(\frac{(z^2-1)^2-4z^2-4zi(z^2-1)}{(z^2+1)^2}\right)dz$$ We keep only the even terms under the integral (this is the "Cauchy principal value" step): $$I=-\pi-4\int_{-1}^1 \frac{(z^2-1)}{(z^2+1)^2}dz \quad (*)$$ $$I=-\pi-4\int_{-1}^1 \left(\frac{1}{z^2+1}-\frac{2}{(z^2+1)^2}\right)dz$$ These integrals are elementary (or tabulated in the case of the last term) and lead to the result $$I=-\pi-4(\pi/2-2\cdot 1/2(1+\pi/2)))=4-\pi$$

Essentially, the only integral to compute is the one marked $(*)$, the rest is just simplification of rational functions. The integral seems to be more trivial than it looks (splitting it as I have is apparently not the most optimal solution, as half of it cancels out). If anyone notices a clever trick to get $\int (x^2-1)/(1+x^2)^2dx=-x/(1+x^2)$ (apart from exclaiming "it's obvious now" when you see the result), let me know in the comments.

Answer 8

Yet another way. Using $u=\tan(x)$, we get $$ \begin{align} \int\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x &=\int\lower{2pt}{\frac{1-\frac{u}{\sqrt{1+u^2}}}{1+\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}}\\ &=\int\left(1-\raise{2pt}{\frac{u}{\sqrt{1+u^2}}}\right)^{\!\!2}\,\mathrm{d}u\\ &=\int\left(2-\raise{2pt}{\frac{2u}{\sqrt{1+u^2}}}-\frac1{1+u^2}\right)\,\mathrm{d}u\\[6pt] &=2u-2\sqrt{1+u^2}-\arctan(u)+C\\[14pt] &=2\tan(x)-2\sec(x)-x+C \end{align} $$ Therefore, $$ \begin{align} \int_0^\pi\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x &=(2-\pi)-(-2)\\[6pt] &=4-\pi \end{align} $$

Answer 9

$\big[$Another (standard) method$\big]$

When we have integrals of the form: $\displaystyle \int \frac{dx}{a+b\sin (x) + c\cos (x)} $ we can perform the substitution: $t=\tan (x/2)$. Then, we have: $\displaystyle \sin(x) = \frac{2t}{1+t^2} $ , $\displaystyle \cos (x) = \frac{1-t^2}{1+t^2} $ , $\displaystyle dx= \frac{2dt}{1+t^2}$.

First we must remove the sine term from the numerator.

$\displaystyle \int_0^{\pi} \frac{1-\sin(x) -1 + 1}{\sin(x) + 1} dx = \int_0^{\pi} \frac{2}{\sin(x)+1} -1 dx$

Now we substitute. Also @$x=0 \rightarrow t=0$ and @$x=\pi \rightarrow t=\infty$. Thus:

$\displaystyle \int_0^{\infty} \Bigg( \frac{2}{\frac{2t}{1+t^2} +1} -1 \Bigg) \cdot \frac{2}{1+t^2} dt = \int_0^{\infty} \frac{4}{2t+1+t^2} - \frac{2}{1+t^2} dt = $

$\displaystyle = \int_0^{\infty} \frac{4}{(t+1)^2} dt - 2\tan^{-1}(t)\big|_0^{\infty} = 4 \left[ -\frac{1}{t+1} \right]_0^{\infty} -2\frac{\pi}{2} = -4 \left( \frac{1}{\infty} - \frac{1}{0+1} \right) -\pi = 4-\pi $

Answer 10

Substitute $t=\sin x:$

$$\begin{align}\int_0^\pi\frac{1-\sin x}{1+\sin x}\mathrm dx&=-\pi+4\int_0^1\frac{\mathrm dt}{(1+t)^\frac32(1-t)^\frac12}\\&\overset{v=\frac{1-t}{1+t}}{=}-\pi+2\int_0^1\frac{\mathrm dv}{\sqrt v}\\&=4-\pi\end{align}$$

Answer 11

With a slight modification to the typical tangent half-angle substitution, the partial fraction expansion can be traded for a more immediate one.

$$\begin{align*} & \int_0^\pi \frac{1-\sin x}{1+\sin x} \, dx \\ &= 2 \int_0^\tfrac\pi2 \frac{1-\sin x}{1+\sin x} \, dx & x\to\pi-x\text{ for }x\in\left[\frac\pi2,\pi\right] \\ &= 4 \int_0^1 \frac{y^2}{1+y^2} \, dy & y=\tan\left(\frac\pi4-\frac x2\right) \\ &= 4 \int_0^1 \left(1-\frac1{1+y^2}\right) \, dy \\ &= 4(1-\arctan1)=\boxed{4-\pi} \end{align*}$$

This is also equivalent to enforcing $u=\dfrac{1-y}{1+y}$ in OP's integral.

Answer 12

$$ \begin{aligned} \int_0^\pi \frac{1-\sin x}{\sin x+1} d x&=2 \int_0^{\frac{\pi}{2}} \frac{1-\sin x}{\sin x+1} d x \\ & =2 \int_0^{\frac{\pi}{2}} \frac{1-\cos x}{\cos x+1} d x \\ & =2 \int_0^{\frac{\pi}{2}} \frac{1-2 \cos x+1-\sin ^2 x}{\sin ^2 x} d x \\ & =2\left(2 \int_0^{\frac{\pi}{2}} \csc ^2 x d x-2 \int_0^{\frac{\pi}{2}} \csc x \cot x d x-\frac{\pi}{2}\right) \\ & =4-\pi \end{aligned}$$

Answer 13

$$\begin{align}\int_0^\pi\frac{1-\sin x}{1+\sin x}\mathrm dx&\overset{(*)}{=}-\pi+4\int_0^\frac\pi2\frac{\mathrm dx}{1+\sin x}\\&=-\pi+4\int_0^\frac\pi2\frac{1-\sin x}{\cos^2x}\mathrm dx\\&=-\pi+4[\tan x-\sec x]_0^\frac\pi2\\&\overset{v=\tan x}{=}-\pi+4+4\lim_{v\to\infty}\frac1{\sqrt{v^2+1}+v}\\&=4-\pi\end{align}$$

Explanation of $(*)$:

$$\int_0^{2a}f(x)\mathrm dx=\int_0^af(x)+f(2a-x)\mathrm dx$$

Answer 14

Substitute $v=\frac1{1+\sin x}$:

$$\begin{align}\int_0^\pi\frac{1-\sin x}{1+\sin x}\mathrm dx&=2 \int_0^\frac\pi2\frac{1-\sin x}{1+\sin x}\mathrm dx\\&=2\int_\frac12^1\frac{\sqrt{2v-1}}v\mathrm dv\\&=4\sqrt{2v-1}-2\sin^{-1}\left(1-\frac1v\right)\Bigg|_\frac12^1\\&=4-\pi\end{align}$$

Answer 15

I'll sugest more tricky approach

First of all :

$ 1 \pm \sin x = \overbrace{ \sin ^2 \frac{x}{2} + \cos^2 \frac{x}{2} }^1 \pm 2 \sin \frac{x}{2}\cdot \cos \frac{x}{2} = (\sin \frac{x}{2} \pm \cos \frac{x}{2} )^2$

Second observation

$\sin \frac{x}{2} + \cos \frac{x}{2} = \sqrt{2}\cos ( \frac{x}{2} - \frac{\pi}{4} ) $

$\sin \frac{x}{2} - \cos \frac{x}{2} = \sqrt{2}\sin ( \frac{x}{2} - \frac{\pi}{4} ) $

Thus

$ \displaystyle \int^{\pi}_0\frac{1-\sin (x)}{\sin (x)+1}dx = \int^{\pi}_0 \left ( \frac{\sin \frac{x}{2} - \cos \frac{x}{2} }{\sin \frac{x}{2} + \cos \frac{x}{2} } \right )^2 dx = \int^{\pi}_0 \left ( \frac{\sqrt{2}\sin ( \frac{x}{2} - \frac{\pi}{4}) }{\sqrt{2}\cos ( \frac{x}{2} - \frac{\pi}{4})} \right )^2 dx = $

$ = \displaystyle \int_{0}^{\pi} \tan^2\left(\frac{x}{2} - \frac{\pi}{4}\right) dx = \int_{0}^{\pi} \left[\sec^2\left(\frac{x}{2} - \frac{\pi}{4}\right) - 1\right] dx $

$\displaystyle = \int_{0}^{\pi} \sec^2\left(\frac{x}{2} - \frac{\pi}{4}\right) dx - \int_{0}^{\pi} 1 \, dx$

$\displaystyle = 2 \tan\left(\frac{x}{2} - \frac{\pi}{4}\right) \Big|_{0}^{\pi} - x \Big|_{0}^{\pi} = 2\left[\tan\left(\frac{\pi}{2} - \frac{\pi}{4}\right) - \tan\left(-\frac{\pi}{4}\right)\right] - \pi $

$= \displaystyle 2\left[\tan\left(\frac{\pi}{4}\right) - (-1)\right] - \pi \\ = 2(1 + 1) - \pi \\ = 4 - \pi$

Answer 16

Using $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$,

$$I = \int_{0}^{\pi}\frac{2i - e^{ix} + e^{-ix}}{e^{ix} - e^{-ix}+2i} ~dx$$

Let $u = e^{ix} \implies x = -i\log(u) \implies dx = \frac{-i}{u} ~du$.

$$I = \int_{1}^{-1}\frac{2i - u + \frac{1}{u}}{u - \frac{1}{u} + 2i}\frac{-i}{u} ~du$$ $$= -i\int_{-1}^{1}\frac{u^2 - 2iu - 1}{u(u^2 + 2iu - 1)} ~du$$

But the integrand is now discontinuous at $u = 0$, so let's split it up.

$$I = -i\int_{-1}^{0}\frac{u^2 - 2iu - 1}{u(u^2 + 2iu - 1)} ~du -i\int_{0}^{1}\frac{u^2 - 2iu - 1}{u(u^2 + 2iu - 1)} ~du$$

Let $v = -u$ in the first integral, and $v = u$ in the second.

$$I = -i\int_{1}^{0}\frac{(-v)^2 - 2i(-v) - 1}{-v((-v)^2 + 2i(-v) - 1)} (-dv) -i\int_{0}^{1}\frac{v^2 - 2iv - 1}{v(v^2 + 2iv - 1)} ~dv$$ $$= i\int_{0}^{1}\frac{v^2 + 2iv - 1}{v(v^2 - 2iv - 1)} ~dv -i\int_{0}^{1}\frac{v^2 - 2iv - 1}{v(v^2 + 2iv - 1)} ~dv$$ $$= i\int_{0}^{1}\frac{(v^2 + 2iv - 1)^2 - (v^2 - 2iv - 1)^2}{v(v^2 - 2iv - 1)(v^2 + 2iv - 1)} ~dv$$ $$= i\int_{0}^{1}\frac{(v+i)^4 - (v-i)^4}{v(v - i)^2(v + i)^2} ~dv$$ $$= i\int_{0}^{1}\frac{(v^4 + 4iv^3 - 6v^2 - 4iv + 1) - (v^4 - 4iv^3 - 6v^2 + 4iv + 1)}{v(v^2 + 1)^2} ~dv$$ $$= i\int_{0}^{1}\frac{8iv^3 - 8iv}{v(v^2 + 1)^2} ~dv$$ $$= -8\int_{0}^{1}\frac{v^2 - 1}{(v^2 + 1)^2} ~dv$$

This eliminates the $i$'s, but let's bring them back to do partial fraction decomposition.

$$I = -8\int_{0}^{1} \left(\frac{1}{2(v+i)^2} + \frac{1}{2(v-i)^2}\right) ~dv$$ $$= -4\int_{0}^{1} \frac{1}{(v+i)^2} ~dv - 4\int_{0}^{1} \frac{1}{(v-i)^2} ~dv$$

In the first integral, let $w = v + i$. In the second integral, let $w = v - i$.

$$I = -4\int_{i}^{1+i} \frac{1}{w^2} ~dw - 4\int_{-i}^{1-i} \frac{1}{w^2} ~dw$$ $$= -4[-\frac{1}{w}]_{i}^{1+i} - 4[-\frac{1}{w}]_{-i}^{1-i}$$ $$= 4\left(\frac{1}{i+1} - \frac{1}{i}\right) + 4\left(\frac{1}{1-i} - \frac{1}{-i} \right)$$ $$= \frac{4(1-i)}{2} - \frac{4}{i} + \frac{4(1+i)}{2} + \frac{4}{i}$$ $$= 2 - 2i + 2 + 2i$$ $$= \boxed{4}$$