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

Hint: 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.