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Best answer
57 votes
57 votes
There is a mod term in the given integral. So, first we have to remove that. We know that $x$ is always positive here and $\sin x$ is positive from 0 to $\pi$. From $\pi$ to $2\pi$, $x$ is positive while $\sin x$ changes sign. So, we can write

$\int_{0}^{2\pi} \mid x \sin x \mid dx = \int_{0}^{\pi} x \sin x dx + \left(-\int_{\pi}^{2\pi} x \sin x dx \right)$

$\qquad\qquad\qquad\quad\;= \int_{0}^{\pi} x \sin x dx -\int_{\pi}^{2\pi} x \sin x dx.$

$\int_{0}^{\pi}{udv} = uv - \int_{0}^{\pi}{vdu}$

$ \text{Here }u = x, du = dx, dv = \sin x dx, \text{ so } v = -\cos x$

$ \therefore \int_{0}^{\pi} x \sin x dx = \left [ -x\cos x\right]_0^{\pi} + \int_{0}^{\pi} \cos x dx$

$= \pi + \left[\sin x\right]_0^{\pi}\\ =\pi$

$\text{Now, } \int_{\pi}^{2\pi} x \sin x = \left [ -x\cos x\right]_{\pi}^{2\pi} + \int_{\pi}^{2\pi} \cos x dx$

$= -3\pi + \left[\sin x\right]_{\pi}^{2\pi}\\ = -3 \pi$

So, given integral = $\pi - \left(-3\pi\right) = 4\pi$

So, k = 4.
edited by
19 votes
19 votes

Some important things, we should know

  • ${\color{Red} {\cos 0 = 1,\cos \pi = -1,\cos 2\pi = 1,\dots}}$
  • ${\color{Blue}{ \text{In general}\: \cos n\pi = (-1)^{n}\: \text{where}\: n=0,1,2,\dots}}$
  • ${\color{Magenta} {\sin 0 = 0,\sin \pi = 0,\sin 2\pi = 0,\dots}}$
  • ${\color{Green} {\text{In general}\: \sin n\pi = 0\: \text{where}\: n=0,1,2,\dots}}$
  • ${\color{Orange} {\sin(\pi-x)=\sin x,\sin (2\pi-x)=-\sin x,\sin(3\pi-x)=\sin x}}$
  • ${\color{Teal} {\text{In general}\: \sin (n\pi-x)=(-1)^{n+1}\sin x\: \text{ where} \: n=0,1,2,\dots}}$
  •  ${\color{Orchid} {\cos(\pi-x)=-\cos x,\cos (2\pi-x)=\cos x,\cos (3\pi-x)=-\cos x}}$
  • ${\color{purple} {\text{ In general}\: \cos (n\pi-x)=(-1)^{n}\cos x \: \text{where}\: n=0,1,2,\dots}}$                                             

$f(x)=\sin x$

$f(x)=\cos x$

Visualization: 

_________________________________________________________________
Let $I = \displaystyle{}\int_{0}^{2\pi} \mid x \sin x \mid {\mathrm{d} x} = k\pi\rightarrow(1)$

Now, we should break the limits and open the 'modulus function'.

$I = \displaystyle{}\int_{0}^{\pi} x \sin x \: {\mathrm{d} x} - \displaystyle{}\int_{\pi}^{2\pi} x \sin x\:  {\mathrm{d} x}$

Now, $I = \left[-x \cos x + \sin x \right]_{0}^{\pi} - \left[-x \cos x + \sin x \right]_{\pi}^{2\pi}$

$\implies I = \left[ \left(- \pi \cos \pi + \sin \pi \right) - \left(-0 \cos 0 + \sin 0 \right) \right] - \left[ \left(- 2\pi \cos 2\pi + \sin 2\pi \right) - \left(- \pi \cos \pi + \sin \pi \right) \right]$

$\implies I = \left[ \left(- \pi (-1) + 0 \right) - \left( 0 + 0 \right) \right] - \left[ \left(- 2\pi (1)   + 0  \right) - \left(- \pi (-1) + 0 \right) \right]$

$\implies I = \pi  + 2\pi + \pi $

$\implies I = 4\pi = k\pi\implies k = 4$

So, the correct answer is $k  = 4.$

edited by
13 votes
13 votes
No need to apply integration by parts,it will be time consuming,

Appy the formula $\int_{0}^{\pi }(\pi-x)sinxdx+\int_{\pi}^{2\pi}(2\pi+\pi -x)sinxdx$

now $I=\int_{0}^{\pi }(\pi-x)sinxdx$

$2I=\pi\int_{0}^{\pi}sinxdx=\pi$

similarly from 2nd part we can get $3\pi$ so total $4\pi$ is the answer
5 votes
5 votes

ILATE PROPERTY IS USEFULL TO DO SUCH TYPE OF QUESTION

BUT ONE ANOTHER method is also important and fast when function is given odd function because both linear algebric and sin function is odd so their product is also odd function and period of sin is \pi and mod function converts them into positive so 

\int_{0}^{\pi }xsinx   gives value  \pi   and   \int_{\pi }^{2\pi } xsinxdx  gives value -3\pi but due to mod function it is gives\left | -3\pi \right |  that is equal to 3\pi    so final answer will be \pi +3\pi = 4\pi

Answer:

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