Question

The value of $\int\limits_{0}^{2x}{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$ where [t] denotes the greatest integer function, is:
$\left( a \right)-2\pi$
$\left( b \right)\pi$
$\left( c \right)-\pi$
$\left( d \right)2\pi$

Answer 1

We have to find the integral of $\left[ \sin 2x\left( 1+\cos 3x \right) \right]$ and we will start by splitting the limit $\left[ 0,2\pi \right]$ as $\left[ 0,\pi \right]$ and $\left[ \pi ,2\pi \right].$ Then we will name the integral $\int\limits_{0}^{\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$ as ${{I}_{1}}$ and ${{I}_{2}}=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}.$ Then we will simplify ${{I}_{2}}$ and we get ${{I}_{2}}$ as $\int\limits_{0}^{\pi }{\left[ -\sin 2x\left( 1+\cos 3x \right) \right]dx}$ and then we use [x] + [– x] = – 1 to simplify and get the integral to get the solution.

Complete step-by-step answer:
We are asked to find the integral of $\left[ \sin 2x\left( 1+\cos 3x \right) \right]$ on the interval from 0 to $2\pi .$ We will firstly split our interval $\left[ 0,2\pi \right]$ as $\left[ 0,\pi \right]$ and $\left[ \pi ,2\pi \right].$ So,
$I=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$
Becomes
$I=\int\limits_{0}^{\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}+\int\limits_{\pi }^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$
Now, let ${{I}_{1}}=\int\limits_{0}^{\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$ and ${{I}_{2}}=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}.$
So, we have
$I={{I}_{1}}+{{I}_{2}}$
First, we will simplify ${{I}_{2}}.$
We have,
${{I}_{2}}=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$
To solve this integral, we will substitute x as $2\pi -t.$
$\Rightarrow x=2\pi -t$
Now, dx will become
$\Rightarrow dx=0-dt$
$\Rightarrow dx=-dt$
Now our earlier limit was $x=\pi$ to $x=2\pi .$ Now, they change in t as,
$t=2\pi -x$
This means, $x=\pi$ becomes $t=\pi .$
While, $x=2\pi$ becomes t = 0. Hence, our integral becomes
${{I}_{2}}=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}$
$\Rightarrow {{I}_{2}}=-\int\limits_{\pi }^{0}{\left[ \sin 2\left( 2\pi -t \right)\left( 1+\cos 3\left( 2\pi -t \right) \right) \right]dt}$
Simplifying, we get,
$\Rightarrow {{I}_{2}}=-\int\limits_{\pi }^{0}{\left[ \sin \left( 4\pi -2t \right)\left( 1+\cos \left( 6\pi -3t \right) \right) \right]dt}$
Now as we know that $\sin \left( 2\pi -\theta \right)=\sin \left( -\theta \right)$ and $\cos \left( 2n\pi -\theta \right)=\cos \left( \theta \right).$ So, we get,
$\Rightarrow {{I}_{2}}=-\int\limits_{\pi }^{0}{\left[ \sin \left( -2t \right)\left( 1+\cos \left( 3t \right) \right) \right]dt}$
Now, sin (– x) = – sin x. So,
$\Rightarrow {{I}_{2}}=-\int\limits_{\pi }^{0}{\left[ -\sin \left( 2t \right)\left( 1+\cos \left( 3t \right) \right) \right]dt}$
Now changing t as x, we get,
$\Rightarrow {{I}_{2}}=-\int\limits_{\pi }^{0}{\left[ -\sin 2x\left( 1+\cos 3x \right) \right]dx}$
We know that, $-\int\limits_{a}^{b}{dx}=\int\limits_{b}^{a}{dx}.$ So, we get,
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{\left[ -\sin 2x\left( 1+\cos 3x \right) \right]dx}$
Now, putting this back in the value of I, we get,
$I={{I}_{1}}+{{I}_{2}}$
$I=\int\limits_{0}^{\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}+\int\limits_{0}^{\pi }{\left[ -\sin 2x\left( 1+\cos 3x \right) \right]dx}$
Now, we know that,
$\left[ x \right]+\left[ -x \right]=-1\text{ }\forall x\notin I$
We will take, $x=\sin 2x\left( 1+\cos 3x \right),$ we get,
$I=\int\limits_{0}^{\pi }{\left[ x \right]dx}+\int\limits_{0}^{\pi }{\left[ -x \right]dx}$
$I=\int\limits_{0}^{\pi }{\left( \left[ x \right]+\left[ -x \right] \right)dx}$
So, using $\left[ x \right]+\left[ -x \right]=-1,$ we get,
$I=\int\limits_{0}^{\pi }{\left( -1 \right)dx}$
$I=\left( -x \right)_{0}^{\pi }$
Putting the limit, we get,
$I=-\pi -\left( -0 \right)$
$I=-\pi$

So, the correct answer is “Option c”.

Note: Remember for $x=2\pi -t,$ we get $dx=-dt$ as we differentiate both the sides, we get,
$dx=d\left( 2\pi -t \right)$
As derivative of the constant is 0.
$\Rightarrow dx=d\left( 2\pi \right)-dt$
So,
$dx=-dt$
Also, when the limit is the same, we can add integral that is why we add ${{I}_{1}}$ and ${{I}_{2}}.$
$I=\int\limits_{0}^{\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]dx}+\int\limits_{0}^{\pi }{\left[ -\sin 2x\left( 1+\cos 3x \right) \right]dx}$
As the limit is the same, so it can be added up.