Question

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

Answer 1

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

& q=\sqrt{x} \\\ & \Rightarrow dq=\dfrac{1}{2\sqrt{x}}dx \\\ \end{aligned}$$ $$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.