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How do we show that?

$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)\tag1$$

$(1)$ it a bit difficult to start with

$$\int_{0}^{\pi}(1+2x)\cdot{\sin(x)[1-\sin^2(x)]\over 1+\cos^2(x)}\mathrm dx\tag2$$

Setting $u=\cos(x)$

$du=-\sin(x)dx$

$$\int_{-1}^{1}(1+2x)\cdot{(u^2)\over 1+u^2}\mathrm du\tag3$$

$$\int_{-1}^{1}(1+2\arccos(u))\cdot{(u^2)\over 1+u^2}\mathrm du\tag4$$

$du=\sec^2(v)dv$

$$\int_{-\pi/4}^{\pi/4}(1+2\arccos(\tan(v)))\tan^2(v)\mathrm dv\tag5$$

$$\int_{-\pi/4}^{\pi/4}\tan^2(v)+2\tan^2(v)\arccos(\tan(v))\mathrm dv=I_1+I_2\tag6$$

$$I_1=\int_{-\pi/4}^{\pi/4}\tan^2(v)\mathrm dv=2-{\pi\over2}\tag7$$

As for $I_2$ I am sure how to do it.

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  • $\begingroup$ your result is not true, i think your integral can not expressed by the known elementary functions $\endgroup$ – Dr. Sonnhard Graubner Jan 22 '17 at 8:31
  • $\begingroup$ yes you have right there was a mistake in my calculations $\endgroup$ – Dr. Sonnhard Graubner Jan 22 '17 at 8:35
  • $\begingroup$ I am not expert but can I ask if integration by parts is possible. taking (1-2x) as the first function. $\endgroup$ – Pushkar Soni Jan 22 '17 at 8:40
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Using the fact that $$ \int_0^\pi xf(\sin x)\,dx=\frac\pi2\int_0^\pi f(\sin x)\,dx $$ and that $\cos^2=1-\sin^2x$ (so that this applies), you get that your integral equals $$ (1+\pi)\int_0^\pi \frac{\sin^3 x}{1+\cos^2x}\,dx $$ Writing $$ \frac{\sin^3 x}{1+\cos^2x}=\frac{(1-\cos^2x)\sin x}{1+\cos^2x}=\frac{2}{1+\cos^2x}\sin x-\sin x $$ we find that your integral equals $$ (1+\pi)\bigl[-2\arctan(\cos x)+\cos x\bigr]_0^\pi=(1+\pi)(\pi-2). $$

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I am not sure this is the easiest (but it is a viable approach). Note that the antiderivative of $v'=\sin^3 x/(1+\cos^2 x)$ is given by $v=\cos x -2 \arctan \cos x$. Thus, we can integrate the problem by parts and obtain $$I = (\cos x -2 \arctan \cos x)(1+2x)\Big|_{x=0}^\pi -2 \underbrace{\int_0^\pi(\cos x -2 \arctan \cos x)dx}_{=I_2}.$$

Now you can observe that $v=\cos x -2 \arctan \cos x$ is antisymmetric around $x=\pi/2$ (because $\cos$ is antisymmetric around $x=\pi/2$ and $\arctan$ is antisymmetric around $x=0$). Thus, the remaining integral vanishes $(I_2=0)$ and we obtain $$I = (\cos x -2 \arctan \cos x)(1+2x)\Big|_{x=0}^\pi= (\pi+1)(\pi-2).$$

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$J=\displaystyle \int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx$

Perform the change of variable $y=\pi-x$,

$\displaystyle J=\int_0^{\pi} (1+2(\pi-x))\dfrac{\sin^3 x}{1+\cos^2 x}dx$

Therefore,

$\begin{align}\displaystyle 2J&=(2+2\pi)\int_0^{\pi}\dfrac{\sin^3 x}{1+\cos^2 x}dx\\ 2J&=-(2+2\pi)\int_0^{\pi}\dfrac{\sin^2 x}{1+\cos^2 x}\text{d}(\cos x)\\ 2J&=-(2+2\pi)\int_0^{\pi}\dfrac{(1-\cos^2 x)}{1+\cos^2 x}\text{d}(\cos x)\\ \end{align}$

Perform the change of variable $y=\cos x$ in the latter integral,

$\begin{align}\displaystyle 2J&=(2+2\pi)\int_{-1}^{1}\dfrac{1-x^2}{1+x^2}dx\\ \displaystyle 2J&=(2+2\pi)\int_{-1}^{1}\dfrac{1}{1+x^2}dx-(2+2\pi)\int_{-1}^{1}\dfrac{x^2}{1+x^2}dx\\ &=(2+2\pi)\Big[\arctan x\Big]_{-1}^{1}-(2+2\pi)\left(\int_{-1}^1 \dfrac{1+x^2}{1+x^2}dx-\int_{-1}^1 \dfrac{1}{1+x^2}dx\right)\\ &=4(2+2\pi)\dfrac{\pi}{4}-2(2+2\pi)\\ &=2(\pi+1)(\pi-2) \end{align}$

Therefore,

$\boxed{J=(\pi+1)(\pi-2)}$

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HINT:

Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

so that $\displaystyle I+I=\int_a^b[f(x)+f(a+b-x)]dx$

Now use $\sin^3x\ dx=-(1-\cos^2x)d(\cos x)$

so replace $\cos x=u$

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  • $\begingroup$ $$\sin(\pi-x)=?,\cos(\pi-x)=?$$ $\endgroup$ – lab bhattacharjee Jan 22 '17 at 9:33
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I am not sure if it is the correct approach,$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx$$ Integrating by parts, taking $(1+2x)$ as first function and then we have:

$$(1+2x)\int_{0}^{\pi}\frac{sin^3x}{1+cos^2x}dx-2\int_{0}^{\pi}(\int_{0}^{v}\frac{sin^3x}{1+cos^2x}dx)dv$$

then for $\int_{0}^{\pi}\frac{sin^3x}{1+cos^2x}dx$ simplify it as: $$\int_{0}^{\pi}\frac{(1-cos^2x)sinx}{1+cos^2x}dx$$ substitute $cosx = t$ and there after it is fairly simple to calculate. by some method of integration.

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  • $\begingroup$ In the second equation the notation is a little messy: a differential is missing and the internal integral should go from $0$ to $x$. $\endgroup$ – N74 Jan 22 '17 at 9:02
  • $\begingroup$ @N74 check I've edited, $0 to x$ where?I haven't substituted anything yet. $\endgroup$ – Pushkar Soni Jan 22 '17 at 9:16
  • $\begingroup$ When you apply integration by parts, the integral you now put in parentheses is an indefinite one, or a definite between 0 and $x$. To avoid confusion I'd also change the 'name' of this variable. $\endgroup$ – N74 Jan 22 '17 at 9:53
  • $\begingroup$ $$(1+2x)\int_{0}^{\pi}\frac{sin^3x}{1+cos^2x}dx-2\int_{0}^{\pi}\left (\int_{0}^{v}\frac{sin^3x}{1+cos^2x}dx\right)dv$$ $\endgroup$ – N74 Jan 22 '17 at 9:56
  • $\begingroup$ @N74 ok i'll edit the same. thanks. $\endgroup$ – Pushkar Soni Jan 22 '17 at 10:01

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