Question

The integration the given function \[\int{\dfrac{\sin \dfrac{5x}{2}}{\sin \dfrac{x}{2}}}dx\] is equal to:
(where $c$ is constant of integration)
(a) $2x+\sin x+2\sin 2x+c$
(b) $x+2\sin x+2\sin 2x+c$
(c) $x+2\sin x+\sin 2x+c$
(d) $2x+\sin x+\sin 2x+c$

Answer 1

Hint: Here the solution we need to Write $\sin \dfrac{5x}{2}=\sin \left( 2x+\dfrac{x}{2} \right)$. Then, we will use the trigonometric formula $\sin \left( A+B \right)=\sin A.\cos B+\cos A.\sin B$ then we will solve the given term and after reducing the equation we will get to the final answer and also in the ending steps we will use the cos x formula $\int{\cos xdx=\sin x}$ then we will get to the final answer.
So here we will start the question

Complete step by step answer:
Let’s start with the given equation then we will get

\[I=\int{\dfrac{\sin \dfrac{5x}{2}}{\sin \dfrac{x}{2}}}dx\]
So, after putting the value we will get,
   \[\Rightarrow I=\int{\dfrac{\sin \left( 2x+\dfrac{x}{2} \right)}{\sin \dfrac{x}{2}}}dx\]
Then we will use the formula $\sin \left( A+B \right)=\sin A.\cos B+\cos A.\sin B$ after using this formula we will get the value.
  \[\Rightarrow I=\int{\left( \dfrac{\sin 2x.\cos \dfrac{x}{2}+\cos 2x.\sin \dfrac{x}{2}}{\sin \dfrac{x}{2}} \right)}dx\]
After solving the equation we will get,
 \[\Rightarrow I=\int{\left( \dfrac{2\sin x.\cos x.\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}+\cos 2x \right)}dx\]
We have to reduce the given question then we will get ,
 \[\begin{align}
  & \Rightarrow I=\int{\left( \dfrac{2.2\sin \dfrac{x}{2}.\cos \dfrac{x}{2}.\cos x.\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}+\cos 2x \right)}dx \\
 & \Rightarrow I=\int{\left( \dfrac{4\sin \dfrac{x}{2}.\cos \dfrac{x}{2}.\cos x.\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}+\cos 2x \right)}dx \\
 & \Rightarrow I=\int{\left( 4\cos \dfrac{x}{2}.\cos x.\cos \dfrac{x}{2}+\cos 2x \right)}dx \\
 & \Rightarrow I=\int{\left( 4{{\cos }^{2}}\dfrac{x}{2}.\cos x+\cos 2x \right)}dx \\
 & \Rightarrow I=\int{\left( 2.2{{\cos }^{2}}\dfrac{x}{2}.\cos x+\cos 2x \right)}dx \\
\end{align}\]
We know the formula $1+\cos 2x=2{{\cos }^{2}}x$ we will put this formula in the above question then we will get the equation as,
 \[\therefore I=\int{\left( 2\left( 1+\cos x \right)\cos x+\cos 2x \right)}dx\]
After reducing the equation then we will get,
\[\begin{align}
  & \Rightarrow I=\int{\left( 2\cos x+2{{\cos }^{2}}x+\cos 2x \right)}dx \\
 & \Rightarrow I=\int{\left( 2\cos x+\left( 1+\cos 2x \right)+\cos 2x \right)}dx \\
 & \Rightarrow I=\int{\left( 2\cos x+1+2\cos 2x \right)}dx \\
 & \Rightarrow I=2\int{\cos x}dx+\int{dx+2\int{\cos 2xdx}} \\
\end{align}\]

After reducing the equation we have to integrate it as We know that $\int{\cos xdx=\sin x}$ and $\int{\cos nxdx=\dfrac{\sin nx}{n}}$ we will apply this in the above equation then we will get,
 $\begin{align}
  & \therefore I=2\sin x+x+2\dfrac{\sin 2x}{2}+c \\
 & \Rightarrow I=2\sin x+x+\sin 2x+c \\
\end{align}$
where $c$ is the constant of integration.
After putting the value of I we will get,
 \[\therefore \int{\dfrac{\sin \dfrac{5x}{2}}{\sin \dfrac{x}{2}}}dx=2\sin x+x+\sin 2x+c\] , where $c$ is the constant of integration.
Hence the correct option is an option (c).

Note:
Here the student must remember the integration of the elementary functions and the formula of trigonometric functions. Mostly the students make mistake in the sign convention like the formula of $\sin \theta $ in the derivation is $\dfrac{d}{dx}\cos x=-\sin x$ there is a negative sign and in integration it is $\int_{{}}^{{}}{\cos xdx=\sin x+c}$ there is no negative sign sometimes we will put the -ve sign-in integration then our answer will get wrong so take care of it.