GATE CSE 2014 | Set 3 | Question: 6

Question

If $\int \limits_0^{2 \pi} |x \: \sin x| dx=k\pi$, then the value of $k$ is equal to ______.

Answer 1

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.

Comments

Can you explain how could you determinal -3pi

I am getting pi only

---

@Sona Barman
[sinx−xcosx]0 to 2π =(sin2π−2πcos2π)−(sinπ−πcosπ)=(−2π)−(π)=−3π.

---

All Silver tea cups 

Answer 2

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 $k  = 4.$

Comments

@Lakshman Patel RJIT This is ultra pro max explanation 😍. Thanks !!

---

Bro, hats off for your explanation ❤️

Answer 3

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

Comments

please elaborate ur method

---

Awesome. Thank You.

---

$f(x)=sinx$

 

$f(x)=cosx$

Answer 4

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  and mod function converts them into positive so 

   gives value     and     gives value  but due to mod function it is gives  that is equal to 3    so final answer will be 

Answer 5

Split in two intervals: 0 to π and π to 2kπ as sin x is negative in π to 2π.

Then use Integration By Parts with ILATE sequence.

Answer is:   4

Comments

We can use the property of definite integral and apply the formula for limits and solve instead of byparts method too

Answer 6

Answer: 4