Question

What is the integral of \[\left( \cos \left( {{x}^{\dfrac{1}{2}}} \right) \right)\]?

Answer 1

Hint: In order to find the integral of \[\left( \cos \left( {{x}^{\dfrac{1}{2}}} \right) \right)\], we will be integrating the given function \[\left( \cos \left( {{x}^{\dfrac{1}{2}}} \right) \right)\] by parts. The general formula for integration by parts is \[\int uv\grave{\ }=uv-\int u\grave{\ }v\]. We will be dividing the given function into parts and then we will integrate them separately. Firstly, we will be integrating \[{{x}^{\dfrac{1}{2}}}\] and then the entire function together.

Complete step-by-step answer:
Now let us have a brief regarding integration by parts. In order to solve using integration by parts, we have to choose \[u\] and \[v\]. And then we have to differentiate \[u\] and then integrate \[v\]. And upon substituting the values in the general formula \[\int uv\grave{\ }=uv-\int u\grave{\ }v\], we obtain the integral.
Now let us find out the integral of the given function \[\left( \cos \left( {{x}^{\dfrac{1}{2}}} \right) \right)\].
\[{{x}^{\dfrac{1}{2}}}\] can be expressed as \[\sqrt{x}\].
Let us consider \[q=\sqrt{x}\]. Now let us differentiate \[q=\sqrt{x}\].
The differentiating rule of \[\sqrt{x}=\dfrac{1}{2\sqrt{x}}\].
So upon differentiating the function, we obtain
\[\begin{align}
  & q=\sqrt{x} \\
 & \Rightarrow dq=\dfrac{1}{2\sqrt{x}}dx \\
\end{align}\]
\[dx=\text{2q dq}\].
\[\therefore \] The integral would be \[2\int q\cos qdq\]
Now let us apply the integration by parts rule i.e. \[\int uv\grave{\ }=uv-\int u\grave{\ }v\]
From our function we have, \[u=q\], \[u\grave{\ }=1\], \[v\grave{\ }=\cos q\] and \[v=\sin q\]
So we have,
\[2\left( q\sin q-\int \sin qdq \right)\]
Upon solving this, we get
\[2\left( q\sin q+\cos q+C \right)\]
Since, earlier we have assumed that \[q=\sqrt{x}\], we will be substituting the value and we get,
\[2\left( \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}+C \right)\]
\[\therefore \] The integral of \[\left( \cos \left( {{x}^{\dfrac{1}{2}}} \right) \right)\] is \[2\left( \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}+C \right)\].

Note: Integration by parts or the partial integration is used in order to find the integral of a product of functions in terms of integral of the product and the derivative and the antiderivative. We have to choose \[u\] such that it would be simpler to differentiate it.