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

Question

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 !

Answer 1

Note this integral can be solved analytically.

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

Answer 2

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 $[0,\pi/4]$.

Answer 3

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:

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

Answer 4

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.

Answer 5

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 $