# Prove $\int_0^{\pi/4} \frac{x \cos 2x}{1-\sin x} \mathrm dx > \frac{1}{4}$

How can I prove this inequality : $$\int_0^{\frac{\pi}{4}} \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x > \frac{1}{4}.$$

It was given by my teacher as a challenge but I can't figure out that how to solve it. For indefinite integral, Usual substitution fails and the product $$x\cos(2x)$$ hints integration by part , but it made it only uglier and hence unsuccessful. Please give a hint or an insightful answer through which it can be easily proven.

Thanks !

• @user1157207 Thanks, but I haven't learnt them yet. Commented Apr 24, 2023 at 8:17
• @QuitMSE The question wasn't given for exams or credits. Commented Apr 24, 2023 at 10:21
• Did you try using the (b-a) int.[ f(a+(b-a)x)] dx formula? It simplifies a lot but I'm getting stuck on furter steps. See if you have any luck. This question is really interesting been hooked on this for about an hour now. Sorry for bad formatting. I'm new to this network Commented Apr 24, 2023 at 11:05
• @An_Elephant It's a trick to reduce the limits of the integral from $a$ to $b$ to $0$ to $1$. $\int_{a}^{b} f(x) dx= (b-a) \int_{0}^{1} f(((b-a)x+a) dx$ Commented Apr 24, 2023 at 13:50
• @An_Elephant No, that's a different one. I never got an opportunity to use that formula in any of the problems so I was sceptic about it. Anyway it's solved now. Good question there is a difference of only 0.02 b/w them Commented Apr 24, 2023 at 19:35

Note this integral can be solved analytically.

\begin{align} \int_0^{\frac{\pi}4} \frac{x \cos(2x)}{1-\sin x} dx&=\int_0^{\frac{\pi}4} \frac{x \cos(2x) (1+\sin x)}{\cos^2 x} dx=\int_0^{\frac{\pi}4} x \cos(2x) (1+\sin x) d\tan x\\ \\ &=-\int_0^{\frac{\pi}4} \cos(2x) (1+\sin x) \tan x~dx\\ &~~~~+2\int_0^{\frac{\pi}4} x\sin(2x) (1+\sin x) \tan x~dx\\ &~~~~-\int_0^{\frac{\pi}4} x\cos(2x) \cos x \tan x~dx=-I_1+I_2-I_3\\ \\ I_1&=\int_0^{\frac{\pi}4} \cos(2x) (1+\sin x) \tan x~dx=^{t=\sin x}=\int_0^{\frac{\sqrt2}2} (1-2t^2) \frac{(1+t)t}{1-t^2}~dt\\ \\ &=\int_0^{\frac{\sqrt2}2} \frac{t(1-2t^2)}{1-t}~dt=\frac{1}2+\frac{2\sqrt2}{3}+\ln\left(\frac{\sqrt2-1}{\sqrt2} \right)\\ \\ I_2&=2\int_0^{\frac{\pi}4} x\sin(2x) (1+\sin x) \tan x~dx=4\int_0^{\frac{\pi}4} x\sin^2x (1+\sin x) ~dx\\ \\ &=\frac{1}2+\frac{13\sqrt2}{9}-\frac{3+5\sqrt2}{12}\pi+\frac{1}{16}\pi^2\\ \\ I_3&=\int_0^{\frac{\pi}4} x\cos(2x) \cos x \tan x~dx=\int_0^{\frac{\pi}4} x\cos(2x) \sin x~dx=\frac{-8+3\pi}{18\sqrt2} \end{align}

• I didn't thought that it could be evaluated. Anyway, can you please expand to show the inequality (without using calculator) and complete the answer? Thanks ! Commented Apr 24, 2023 at 14:39
• @An_Elephant Isn't this answer done? How is it incomplete? Commented Apr 24, 2023 at 15:25
• @Accelerator Hi, The answer evaluates the integral. However, it don't shows that how the value of integral is greater than $\frac{1}{4}$ . I mean , that in last, it don't use calculus(or other math methods) to show the inequality. One has to rely on calculator to confirm the inequality. Commented Apr 24, 2023 at 15:41
• @An_Elephant This is very unoptimal but proves that it can be done by hand (theoretically). The closed form reduces to $$I=-\ln\left(1-\frac1{\sqrt2}\right)+\sqrt2-\frac\pi{\sqrt2}+\frac{\pi^2}{16}-\frac\pi4.$$ The inequalities $223/71<\pi<355/113$ and $1.414<\sqrt2$ can all be shown analytically, and using the Maclaurin expansion of $\ln(1-x)$ we obtain after much simplification $$I>\sum_{n=1}^{15}\frac{0.707^n}n-\frac{786628308457}{805461062000}.$$ It is theoretically possible to prove by hand that this is greater than $1/4$, but this would really be a waste of time more than anything else. Commented Apr 25, 2023 at 22:55
• @TheSimpliFire Okay thanks, I got it. I had initially thought that it's solution would be like the integrand $< f(x)$ for some function $f$ and integral of $f$ would be $1/4$ .Anyway, thanks ! Commented Apr 26, 2023 at 0:18

We first show that for $$x>0$$ $$x>\sin(x)+\frac{\sin(x)^3}{6}\qquad(\ast)$$ (note that $$\arcsin(t)=t+\frac{t^3}{6}+O(t^5)$$).

Indeed $$f(x):=x -\sin(x)-\frac{\sin(x)^3}{6}$$ is such that $$f(0)=0$$ and $$f$$ is increasing for $$x>0$$: $$f'(x)=1-\cos(x)-\frac{\sin(x)^2\cos(x)}{2}=\frac{(2+\cos(x))(1-\cos(x))^2}{2}\geq 0.$$

Therefore, by $$(\ast)$$, $$I:=\int_0^{\pi/4} \frac{x \cos(2x)}{1-\sin x} \, dx > \int_0^{\pi/4} \frac{\big(\sin(x)+\frac{\sin(x)^3}{6}\big) \cos(2x)}{1-\sin(x)} \,dx.$$ It remains to show that the trigonometric integral on the right, which can be done with standard techniques, is greater than $$1/4$$. Please fill the details.

P.S. I verified that the inequality $$(\ast)$$ works pretty well: by WA, $$I_1:=\int_0^{\pi/4} \frac{\sin(x) \cos(2x)}{1-\sin(x)} \,dx=\frac{3}{2}-2\sqrt{2}+\frac{\pi}{2},$$ and by WA, $$I_2:=\int_0^{\pi/4} \frac{\sin(x)^3 \cos(2x)}{1-\sin(x)} \,dx=\frac{19}{12}-\frac{7\sqrt{2}}{3}+\frac{9\pi}{16}.$$ Hence $$0.25217\approx I>I_1+\frac{I_2}{6}=\frac{127}{72}-\frac{43\sqrt{2}}{18}+\frac{19\pi}{32}\approx0.25081>\frac{1}{4}.$$ Also comparing the graphs of the two integrands, we find that they are VERY close over the interval $$[0,\pi/4]$$.

• But can't we show the inequality using calculus only (without calculator or approximations) ? Thanks . Commented Apr 24, 2023 at 15:05
• In order to show that $\frac{127}{72}-\frac{43\sqrt{2}}{18}+\frac{19\pi}{32}>\frac{1}{4}$ without calculator you can use a rational lower bound of $\pi$ (such us $223/71$ known to Archimedes). Then you are left to compare a rational number with $\sqrt{2}$ which can be done by squaring. Commented Apr 24, 2023 at 16:06
• But I was looking for something non-calculating and non-numerical. Anyway, thanks ! Commented Apr 24, 2023 at 16:11
• @RobertZ I found that my solution is the same as yours (fortunately, I did not post it before I read all solutions here. Then I found your solution.). First, I used $x \ge \sin x$, but $\int_0^{\pi/4} \frac{\sin x \cos 2x}{1-\sin x} \mathrm dx \approx 0.2423692020$ which is not enough. Then I tried $x \ge \sin x + \frac{\sin^3 x}{6}$ and it works. Commented May 29, 2023 at 13:49

We can start by using integration by parts, where we let $$u=x$$ and $$\mathrm{d}v=\frac{\cos(2x)}{1-\sin(x)}\mathrm{d}x$$. Then, $$\mathrm{d}u=\mathrm{d}x$$ and we can let $$v$$ be the antiderivative of $$\frac{\cos(2x)}{1-\sin(x)}$$:

$$v = \int \frac{\cos(2x)}{1-\sin(x)}\mathrm{d}x$$

We can solve this integral by substituting $$u=\sin(x)-1$$:

\begin{align*} v &= \int \frac{\cos(2x)}{1-\sin(x)}\mathrm{d}x\\ &= -\frac{1}{2}\int \frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{1}{u}\right)\mathrm{d}u\\ &= -\frac{1}{2}\ln|u| + C\\ &= -\frac{1}{2}\ln|1-\sin(x)| + C \end{align*}

Using the limits of integration $$0$$ and $$\frac{\pi}{4}$$, we have:

\begin{align*} \int_0^{\frac{\pi}{4}} \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x &= \left[x\ln|1-\sin(x)|\right]_0^{\frac{\pi}{4}} + \frac{1}{2}\int_0^{\frac{\pi}{4}} \ln|1-\sin(x)|\mathrm{d}x\\ &= \frac{\pi}{4}\ln\left|\frac{1-\sqrt{2}/2}{1+\sqrt{2}/2}\right| + \frac{1}{2}\int_0^{\frac{\pi}{4}} \ln|1-\sin(x)|\mathrm{d}x\\ &= \frac{\pi}{4}\ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| + \frac{1}{2}\int_0^{\frac{\pi}{4}} \ln|\cos^2(x/2)|\mathrm{d}x\\ &= \frac{\pi}{4}\ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| + \int_0^{\frac{\pi}{8}} \ln(\cos(u))\mathrm{d}u\\ &= \frac{\pi}{4}\ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| + \int_0^{\frac{\pi}{8}} \ln(\cos(\frac{\pi}{8}-u))\mathrm{d}u\\ &= \frac{\pi}{4}\ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| + \int_0^{\frac{\pi}{8}} \ln(\sin(u)+\cos(u))\mathrm{d}u\\ &> \frac{\pi}{4}\ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| \end{align*}

The inequality follows from the fact that $$\ln(\sin(u)+\cos(u)) > \ln(1) = 0$$ for $$0 \leq u \leq \frac{\pi}{8}$$.

Now, we just need to show that $$\frac{\pi}{4}\ln\left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right| > \frac{1}{4}$$.

We can simplify the expression inside the logarithm using the difference of squares:

$$\frac{2-\sqrt{2}}{2+\sqrt{2}} = \frac{(2-\sqrt{2})^2}{2^2-(\sqrt{2})^2} = 3-2\sqrt{2}$$

Substituting this back into the integral, we have:

\begin{align*} \int_0^{\frac{\pi}{4}} \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x &> \frac{\pi}{4}\ln\left|3-2\sqrt{2}\right|\\ &= \frac{\pi}{4}\ln(2-\sqrt{2}) - \frac{\pi}{4}\ln 2\\ &= \frac{\pi}{4}\ln(2-\sqrt{2}) - \frac{\ln 2}{4} \end{align*}

We can show that $$\ln(2-\sqrt{2}) > \frac{1}{2}$$ by using the fact that $$e^x > 1+x$$ for all $$x$$, which implies that $$\ln(1+x) > x$$ for all $$x > -1$$. Setting $$x=-\sqrt{2}$$, we have:

$$\ln(2-\sqrt{2}) = \ln(1+(\sqrt{2}-1)) > \sqrt{2}-1 > \frac{1}{2}$$

Therefore, we have:

\begin{align*} \int_0^{\frac{\pi}{4}} \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x &> \frac{\pi}{4}\ln(2-\sqrt{2}) - \frac{\ln 2}{4}\\ &> \frac{\pi}{4}\cdot\frac{1}{2} - \frac{\ln 2}{4}\\ &= \frac{\pi}{8} - \frac{\ln 2}{4}\\ &> \frac{3}{8} - \frac{1}{2}\\ &= \frac{1}{4} \end{align*}

Therefore, we have proven that:

$$\int_0^{\frac{\pi}{4}} \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x > \frac{1}{4}$$

• Good job! I like it. Commented Apr 30, 2023 at 15:29
• Your antiderivative of $\int \frac{\cos(2x)}{1-\sin(x)}\mathrm{d}x$ doesn't seem right. Differentiating back $v$ does not give the original expression. Thanks ! Commented Apr 30, 2023 at 16:54
• Will check and revise. Appreciate! Commented May 1, 2023 at 0:24

In fact, using $$\int \frac{\cos(2x)}{1-\sin(x)} \mathrm{d}x=\int \frac{1-2\sin^2x}{1-\sin(x)} \mathrm{d}x=\int \frac{(2-2\sin^2x)-1}{1-\sin(x)} \mathrm{d}x=2x-2\cos(x)-\int \frac{1}{1-\sin(x)}=2x-2\cos(x)-\frac{2\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}+C$$ and $$\int\frac{\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}\mathrm{d}x=-\frac x2-\ln\bigg(\cos(\frac x2)-\sin(\frac x2)\bigg)+C$$ one has, by IBP, $$\begin{eqnarray} &&\int \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x \\ &=&x\bigg[2x-2\cos(x)-\frac{2\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}\bigg]-\int\bigg[2x-2\cos(x)-\frac{2\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}\bigg]\mathrm{d}x \\ &=&x\bigg[2x-2\cos(x)-\frac{2\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}\bigg]-\bigg[x+x^2+2\ln\bigg(\cos(\frac x2)-\sin(\frac x2)\bigg)-2\sin(x)\bigg] \\ &=&x^2-x+2\sin(x)-2\cos(x)-\frac{2\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}-2\ln\bigg(\cos(\frac x2)-\sin(\frac x2)\bigg) \end{eqnarray}$$ and hence $$\int_0^{\frac{\pi}{4}} \frac{x \cos(2x)}{1-\sin(x)} \mathrm{d}x=x^2-x+2\sin(x)-2\cos(x)-\frac{2\sin(\frac x2)}{\cos(\frac x2)-\sin(\frac x2)}-2\ln\bigg(\cos(\frac x2)-\sin(\frac x2)\bigg)\bigg|_0^{\frac\pi4}\approx0.2521>\frac14.$$ I don't think there is a simple way to show it.

• That's the important part of question :( We cannot show it without using calculator ? Thanks ! Commented Apr 24, 2023 at 15:04

Here's my try (This method is super lengthy and not advised);

we have, $$\displaystyle I=\int_0^{\frac{\pi}{4}} \frac{x\cos(2x)}{1-\sin(x)} \mathrm{d}x$$,

$$\left[\text{using}, \, \cos(2x) = 1 -2\sin^2(x)\right]$$

$$\displaystyle I=\int_0^{\frac{π}{4}} \frac{x(1 -2\sin^2(x))}{1-\sin(x)} \mathrm{d}x$$, which can be split as two integrals, $$I_1$$ and $$I_2$$ ($$I_1 - 2I_2$$),

where, $$\displaystyle I_1 = \int_0^{\frac{π}{4}} \frac{x}{1-\sin(x)} \,\mathrm{d}x$$ and $$\displaystyle I_2=\int_0^{\frac{\pi}{4}} \frac{x\sin^2(x)}{1-\sin(x)} \mathrm{d}x$$

Further, $$\displaystyle I_1= \int_0^{\frac{\pi}{4}} \frac{x}{1-\sin(x)} \,\mathrm{d}x = \int_0^{\frac{\pi}{4}} \frac{x\sec(x)}{\tan(x)-\sec(x)} \,\mathrm{d}x$$ [by multiplying and dividing by $$\sec(x)$$ ]

$$\displaystyle I_1 = \ln(1-\sin(x))- \frac{x}{\tan(x)-\sec(x)}$$

On applying the bounds, we get $$I_1 = \ln(2-\sqrt2) - \ln(2) + \dfrac{\pi}{2^{\frac{3}{2}}(2-\sqrt2)} \approx 0.66817172$$.

Now, $$\displaystyle I_2=\int_0^{\frac{\pi}{4}} \frac{x\sin^2(x)}{1-\sin(x)} \,\mathrm{d}x$$ [that '2' can taken in the end as its a constant], using [$$\cos(x)^2=1-\sin(x)^2$$],

$$\displaystyle I_2 = \int_0^{\frac{\pi}{4}} \frac{x}{1-\sin(x)} \,\mathrm{d}x - \int_0^{\frac{\pi}{4}} \dfrac{x\cos^2(x)}{1-\sin(x)} \,\mathrm{d}x = I_1 - I_3$$,

$$\displaystyle I_3 = \int_0^{\frac{\pi}{4}} \frac{x\cos^2(x)}{1-\sin(x)} \,\mathrm{d}x$$ can be solved by [$$\sin(x)^2=1-\cos(x)^2$$] and applying bounds,

$$I_3 \approx 0.4601715$$

So, $$I = I_1-2(I_1-I_3) = 2(I_3) - (I_1) \approx 2(0.4601715)-(0.66817172) = 0.920343-0.66817172 = 0.25217128 > \dfrac{1}{4}$$.

PS- How was $$I_1$$ solved?

$$\displaystyle I_1 = \int_0^{\frac{\pi}{4}} \frac{x\sec(x)}{\tan(x)-\sec(x)} \,\mathrm{d}x$$, taking '$$x$$' as first function and $$\dfrac{\sec(x)}{\sec(x)+\tan(x)}$$ as seond function, we have integration by parts,

$$(first) \int (second)\, \mathrm{d}x$$ - $$\int (first)' \left[\int (second)\, \mathrm{d}x\right] \, \mathrm{d}x$$, using this we get,

$$\displaystyle I_1 = x \int \frac{\sec(x)}{\tan(x)-\sec(x)} \,\mathrm{d}x - \int (1) \int \frac{\sec(x)}{\tan(x)-\sec(x)} \,\mathrm{d}x\, \mathrm{d}x$$

$$\left[\text{Put }\frac{1}{\sec(x)+\tan(x)}\text{ as }u\text{ and solve, i.e, }\displaystyle\int \frac{\sec(x)}{\sec(x)+\tan(x)}\mathrm{d}x = \dfrac{-1}{\sec(x)+\tan(x)}\right]$$

$$\displaystyle I_1 = x \int \frac{\sec(x)}{\tan(x)-\sec(x)} \,\mathrm{d}x - \int (1) \int \frac{\sec(x)}{\tan(x)-\sec(x)} \,\mathrm{d}x\, \mathrm{d}x$$ (use the above result)

How was $$I_3$$ solved?

$$\displaystyle I_3 = \int_0^{\frac{\pi}{4}} \frac{x\cos(x)^2}{1-\sin(x)} \,\mathrm{d}x$$ ( as mentioned above, use $$\cos(x)^2=1-\sin(x)^2$$ )

$$\displaystyle I_3 = \int \frac{x(1-\sin(x)^2)}{1-\sin(x)} \,\mathrm{d}x = \int \frac{x}{1-\sin(x)} \,\mathrm{d}x - \int \frac{x\sin(x)^2}{1-\sin(x)} \,\mathrm{d}x$$,

here $$I_3=I_1 - I_4$$(say)

$$\displaystyle I_4 = \int \frac{x\sin(x)^2}{1-\sin(x)} \mathrm{d}x$$

• Can you show your steps for $I_3$ ? I am unsure how you calculated $I_1$ using by parts ? Thanks Commented Apr 24, 2023 at 10:36
• yeah ill edit it in Commented Apr 24, 2023 at 10:36
• $I_4$ = $\int \frac{x\sin^2x}{1-\sin x} \mathrm dx$ and not $\int \frac{\sin^2x}{1-\sin x} \mathrm dx$ , right ? Commented Apr 24, 2023 at 11:34
• No , I didn't. Unfortunately, I have known about this exam just now. Commented Apr 28, 2023 at 11:16