Question

How do you find the integral of $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$?

Answer 1

Hint: The given integral $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$ has \[{{x}^{\dfrac{1}{2}}}\] or \[\sqrt{x}\] as an argument to the sine function, which is making it complex. So we will simplify the integral by substituting \[{{x}^{\dfrac{1}{2}}}\] equal to some variable, say $t$. On making this substitution our integral will become simplified. Then we have to use the by-parts method to solve the integral obtained. Finally, we have to back substitute $t$ to \[{{x}^{\dfrac{1}{2}}}\] to get the final integral.

Complete step-by-step answer:
The integral given in the above question is
\[I=\int{\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx}..........(i)\]
As can be seen above, the square root function \[{{x}^{\dfrac{1}{2}}}\] as an argument to the sine function is making the integral complex. So we first have to simplify the above integral by removing the square root function by substituting it to some variable $t$, that is,
$\Rightarrow t={{x}^{\dfrac{1}{2}}}.........(ii)$
Differentiating both sides with respect to $x$, we have
$\Rightarrow \dfrac{dt}{dx}=\dfrac{d\left( {{x}^{\dfrac{1}{2}}} \right)}{dx}$
Now, we know that the differentiation of ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. So the above equation becomes
\[\begin{align}
  & \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \\
 & \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \\
 & \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2{{x}^{\dfrac{1}{2}}}} \\
\end{align}\]
Substituting (ii) in the above equation, we get
\[\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2t}\]
By cross multiplying, we can write
\[\begin{align}
  & \Rightarrow 2tdt=dx \\
 & \Rightarrow dx=2tdt.........(iii) \\
\end{align}\]
Substituting (ii) and (iii) in (i), we get
$\begin{align}
  & \Rightarrow I=\int{\sin t\left( 2tdt \right)} \\
 & \Rightarrow I=\int{2t\sin tdt}......(iv) \\
\end{align}$
Now, we will use the by parts method to solve the above integral. From the by parts method, we know that
$\int{f\left( t \right)g\left( t \right)dt}=f\left( t \right)\int{g\left( t \right)dt}-\int{f'\left( t \right)\left( \int{g\left( t \right)dt} \right)dt}$
Choosing $f\left( t \right)=2t$ and $g\left( t \right)=\sin t$, the integral in (iv) can be written as
\[\begin{align}
  & \Rightarrow I=2t\int{\sin tdt}-\int{\dfrac{d\left( 2t \right)}{dt}\left( \int{\sin tdt} \right)dt} \\
 & \Rightarrow I=2t\int{\sin tdt}-\int{2\left( \int{\sin tdt} \right)dt} \\
\end{align}\]
We know that \[\int{\sin tdt}=-\cos t\]. Putting it in the above integral, we get
\[\begin{align}
  & \Rightarrow I=2t\left( -\cos t \right)-\int{2\left( -\cos t \right)dt} \\
 & \Rightarrow I=-2t\cos t+2\int{\cos tdt} \\
\end{align}\]
Now, we know that \[\int{\cos tdt}=\sin t\]. Putting it above, we get
\[\begin{align}
  & \Rightarrow I=-2t\cos t+2\sin t+C \\
 & \Rightarrow I=2\sin t-2t\cos t+C \\
\end{align}\]
Finally, substituting (ii) in the above equation, we get
\[\begin{align}
  & \Rightarrow I=2\sin {{x}^{\dfrac{1}{2}}}-2{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}}+C \\
 & \Rightarrow I=2\left( \sin {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}} \right)+C \\
\end{align}\]
Hence, the integral of $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$ is equal to \[2\left( \sin {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}} \right)+C\].

Note: Do not forget to add a constant after the integration since we have solved an indefinite integral. Also, while substituting ${{x}^{\dfrac{1}{2}}}=t$, do not replace $dx$ by $dt$ directly. We have to take the differential on both sides of the equation ${{x}^{\dfrac{1}{2}}}=t$ for obtaining $dt$ in terms of $dx$.