$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$

Question

MyAttempt

$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$

I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sign or how to break it in further intervals?

You should try to write $a\sin x+ b\cos x$ as $\sin(x+\alpha)$. — Quang Hoang, Aug 20 '15 at 05:01
This is mechanized in Maple. The answer is $2,\sqrt {2}\sqrt {3}.$ — user64494, Aug 20 '15 at 06:23

score 2 · Answer 1 · answered Aug 20 '15 at 05:08

2

Hint: For any $\theta\in\mathbb{R}$, $\displaystyle\int_\theta^{\pi+\theta}\,\big|\sin(x)\big|\,\text{d}x=\int_0^{\pi}\,\big|\sin(x)\big|\,\text{d}x=2$. Also, $$\sqrt{2}\sin(x)+2\cos(x)=\sqrt{6}\sin\big(x+\arctan(\sqrt{2})\big)\,.$$

answered Aug 20 '15 at 05:08

Batominovski

49,629

juantheron · Accepted Answer · 2015-08-20T05:10:06.967

We Can Write $\displaystyle \sqrt{2}\sin x+2\cos x = \sqrt{\left(\sqrt{2}\right)^2+2^2}\left\{\frac{\sqrt{2}}{\sqrt{6}}\sin x+\frac{2}{\sqrt{6}}\cos x\right\}$

$\displaystyle = \sqrt{6}\left\{\sin x\cdot \frac{1}{\sqrt{3}}+\cos x\cdot \frac{2}{\sqrt{6}}\right\} = \sqrt{6}\left\{\sin x\cdot \cos \alpha+\cos x\cdot \sin \alpha\right\} = \sqrt{6}\cdot \sin (x+\alpha)$

Where $$\displaystyle \sin \alpha = \frac{2}{\sqrt{6}}$$ and $$\displaystyle \cos \alpha = \frac{1}{\sqrt{3}}$$ and $$\displaystyle \tan \alpha = \sqrt{2}$$ . So $\displaystyle \frac{\pi}{4}<\alpha <\frac{\pi}{3}$

So Integral Convert into $$\displaystyle \sqrt{6}\int_{0}^{\pi}\left|\sin (x+\alpha)\right|dx$$

Let $(x+\alpha) = t\;,$ Then $dx = dt$ and Changing Limit, we get

$$\displaystyle I = \sqrt{6}\int_{\alpha}^{\pi+\alpha}|\sin t|dt = \sqrt{6}\int_{\alpha}^{\pi}\sin tdt-\sqrt{6}\int_{\pi}^{\pi+\alpha}\sin tdt $$

So we get $$\displaystyle I = \sqrt{6}\left(1+\cos \alpha\right)+\sqrt{6}\left(-\cos \alpha +1\right) = 2\sqrt{6}$$

score 1 · Answer 3 · answered Aug 20 '15 at 05:04

1

The break point occurs when $\sqrt{2}\sin x+2\cos x=0$, which is at

$$x=\pi-\arctan(\sqrt{2})$$

Thus, the integral is

$$\begin{align} \int_0^{\pi}|\sqrt{2}\sin x+2\cos x|\,dx&=\int_0^{\pi-\arctan(\sqrt{2})}(\sqrt{2}\sin x+2\cos x)\,dx\\\\ &-\int_{\pi-\arctan(\sqrt{2})}^{\pi}(\sqrt{2}\sin x+2\cos x)\,dx \end{align}$$

Can you finish from here?

answered Aug 20 '15 at 05:04

Mark Viola

179,405

Using this break up,i got answer$2\sqrt6-\sqrt2$.Correct answer is $2\sqrt6$.Is something missing in your suggestion.Thanks – user1442 Aug 20 '15 at 05:27
@user1442 I just confirmed my analysis on Wolfram Alpha - the answer using this approach is also $2\sqrt{6}$. – Mark Viola Aug 20 '15 at 05:34
HERE IS THE LINK that supports the analysis. – Mark Viola Aug 20 '15 at 05:35
Yes i rechecked my calculations,This method is giving perfect answer.Thanks – user1442 Aug 20 '15 at 05:42
You're welcome. My pleasure. I thought that it was the answer you were seeking. – Mark Viola Aug 20 '15 at 05:43
What is the reason behind taking break point at $x=\pi-\arctan\sqrt2$ – user1442 Aug 20 '15 at 05:43
1

That is the point for which $\sqrt{2}\sin x+2\cos x=0$. – Mark Viola Aug 20 '15 at 05:43
solving the equation $\sqrt2 \sin x+2\cos x=0$ gives us $x=\arctan(-\sqrt2)?$ – user1442 Aug 20 '15 at 05:45
1

@user1442 The range of the arctangent function is $[-\pi/2,\pi/2]$. We know we are beyond $\pi/2$, correct? So, we need to be a bit careful when finding the solution to that equation. – Mark Viola Aug 20 '15 at 05:52

Claude Leibovici · Answer 4 · 2015-08-20T05:14:32.883

Hint

Consider the function $$f(x)=\sqrt2\sin (x)+2\cos (x)$$ $$f'(x)=\sqrt2\cos(x)-2\sin (x)$$ The function starts being positive, goes throught a maximum and then decreases and cancels somewhere in the interval. So, find the root $a$ such that $f(a)=0$ and now $$I=\int_{0}^{\pi}|\sqrt2\sin (x)+2\cos (x)|\,dx$$ $$I=\int_{0}^{a}\Big(\sqrt2\sin(x)+2\cos (x)\Big)dx-\int_{a}^{\pi}\Big(\sqrt2\sin(x)+2\cos (x)\Big)dx$$ $$I=4 \sin (a)-2 \sqrt{2} \cos (a)$$

score 1 · Answer 5 · answered Aug 22 '15 at 05:54

$\displaystyle A \sin \ x + B \cos \ x = \sqrt{A^2 + B^2}\sin( x + \phi )$, where $\displaystyle \cos \phi = \frac{A}{\sqrt{A^2 + B^2}}$.

In our case $\displaystyle \cos\phi=\frac{1}{\sqrt{3}} \Rightarrow \phi\in\left(0;\frac{\pi}{2}\right)$.

Thus: $\displaystyle \begin{aligned}\int\limits_0^\pi\left|\sqrt{2}\sin x+2\cos x\right|dx&=\sqrt{6}\int\limits_0^\pi\left|\sin(x+\phi)\right|dx=\sqrt{6}\int\limits_0^{\pi-\phi}\sin(x+\phi)dx+\sqrt{6}\int\limits_{\pi-\phi}^\pi(-\sin(x+\phi))dx=\\&=\left.-\sqrt{6}\cos(x+\phi)\right|_0^{\pi-\phi}+\left.\sqrt{6}\cos(x+\phi)\right|_{\pi-\phi}^\pi=-\sqrt{6}\left(\cos\pi-\cos\phi\right)+\sqrt{6}\left(\cos(\pi+\phi)-\cos\pi\right )=\\&=\sqrt{6}+\sqrt{2}-\sqrt{2}+\sqrt{6}=2\sqrt{6}\end{aligned}$

this answer seems to be very similar to your other answer :-) — Math-fun, Aug 22 '15 at 06:56

Harish Chandra Rajpoot · Answer 6 · 2015-08-20T05:39:53.993

0

Let's first find out the break point $$\sqrt 2\sin x+2\cos x=0$$ $$\sqrt 2\sin x=-2\cos x\iff \tan x=-\frac{2}{\sqrt 2}$$ $$x=\tan^{-1}(-\sqrt 2)$$

Now, we have $$\int_{0}^{\pi}\left|\sqrt 2\sin x+2\cos x \right|dx$$ $$=\int_{0}^{\tan^{-1}(-\sqrt 2)}\left|\sqrt 2\sin x+2\cos x \right|dx+\int_{\tan^{-1}(-\sqrt 2)}^{\pi}\left|\sqrt 2\sin x+2\cos x \right|dx$$ $$=\int_{0}^{\tan^{-1}(-\sqrt 2)}(\sqrt 2\sin x+2\cos x)dx-\int_{\tan^{-1}(\sqrt 2)}^{\pi}(\sqrt 2\sin x+2\cos x )dx$$

I hope you can solve further

edited Aug 20 '15 at 05:39

answered Aug 20 '15 at 05:16

Harish Chandra Rajpoot

37,450

This method will not give me $2\sqrt6$ which is the correct answer.Thanks.. – user1442 Aug 20 '15 at 05:37
the root is wrong, it should be $\arctan(-\sqrt{2})+\pi$. – Math-fun Aug 22 '15 at 06:54
Kindly, notice $$\tan^{-1}(-\sqrt 2)=\pi-\tan^{-1}(\sqrt 2)$$ – Harish Chandra Rajpoot Aug 22 '15 at 06:59

$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$

6 Answers6