The indefinite integral of the trigonometric function sine of angle the square root of $x$ should be evaluated with respect to $x$ in this integration problem.

The given function is a function, formed by the composition of a trigonometric function sine and an irrational function. Let’s learn how to find the indefinite integral of the sine of $\sqrt{x}$ with respect to $x$.

There is an integral rule in calculus to find the integration of sine function but there is no rule to find the integral of sine of the square root of $x$ with respect to $x$. The square root function makes the integration tough. So, let us try to eliminate the square root function by the change of variable trick.

Let’s take $y \,=\, \sqrt{x}$

Now, differentiate the equation with respect to $x$ to find the derivative of square root of $x$.

$\implies$ $\dfrac{d}{dx}\,y$ $\,=\,$ $\dfrac{d}{dx}\,\sqrt{x}$

On the left hand side of the equation, it is not possible to find the derivative of $y$ with respect to $x$. So, it is just written as derivative of $y$ with respect to $x$. However, the derivative of square root of $x$ can be evaluated with respect to $x$ on the right hand side of the equation.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{1}{2\sqrt{x}}$

The square root of $x$ is denoted by a variable $y$. So, the $\sqrt{x}$ function can be replaced by $y$ on the right hand side of the equation.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{1}{2y}$

The expressions on both sides of the equation are in ratio form. So, it can be simplified by the cross multiply method.

$\implies$ $2y \times dy$ $\,=\,$ $1 \times dx$

$\implies$ $2ydy$ $\,=\,$ $dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $2ydy$

Now, let us change the integral function in terms of $x$ into the integral function in terms of $y$.

$\implies$ $\displaystyle \int{\sin{\sqrt{x}}}\,dx$ $\,=\,$ $\displaystyle \int{\sin{y}}\,2ydy$

The expression on the right hand side of the equation can be simplified further to write the function in simple form.

$=\,\,$ $\displaystyle \int{\sin{y}}\, \times 2y \times dy$

According to the commutative property, the places of the factors can be changed in the multiplication.

$=\,\,$ $\displaystyle \int{2y \times \sin{y}}\, \times dy$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{\sin{\sqrt{x}}}\,dx$ $\,=\,$ $\displaystyle \int{2y\sin{y}}\,dy$

It expresses that the integral of sine of angle square root of $x$ can be evaluated with respect to $x$ by finding the integral of two times the product of $y$ and sine of angle $y$ with respect to $y$.

Let’s discuss about the function in terms of $y$ firstly. In this function, $2$ is a constant function, $y$ is an identity function and $\sin{y}$ is a trigonometric function. The integral of the product of three functions should be evaluated with respect to $y$. So, the integration by parts can be used to find the product of the functions.

$\displaystyle \int{2 \times y\sin{y}}\,dy$

In the function, $2$ is a number. So, it can be released from the integral operation as per the constant multiple rule of integration.

$=\,\,$ $2 \times \displaystyle \int{y\sin{y}}\,dy$

Now, let us find the indefinite integral of the product of $y$ and trigonometric function $\sin{y}$ with respect to $y$ by the integration by parts.

Let’s take $u = y$ and $dv = \sin{y}\,dy$

Firstly, differentiate the equation $u = y$ with respect to $y$ to find the differential element $du$.

$\implies$ $\dfrac{d}{dy}\,{u} \,=\, \dfrac{d}{dy}\,{y}$

$\implies$ $\dfrac{du}{dy} \,=\, \dfrac{dy}{dy}$

According to the integral rule of a variable, the derivative of $y$ with respect to $y$ is equal to one.

$\implies$ $\dfrac{du}{dy} \,=\, 1$

$\implies$ $du \,=\, 1 \times dy$

$\,\,\,\therefore\,\,\,\,\,\,$ $du \,=\, dy$

Now, integrate both sides of the equation $dv = \sin{y}\,dy$ to find the variable $y$.

$\implies$ $\displaystyle \int{}dv \,=\, \int{\sin{y}}\,dy$

$\implies$ $\displaystyle \int{1}dv \,=\, \int{\sin{y}}\,dy$

According to the integral rule of one, the integral of $1$ with respect $v$ is equal to $v$, and the integral of sine of angle $y$ with respect to $y$ is equal to negative cosine of angle $y$ as per the integral rule of sine function.

$\,\,\,\therefore\,\,\,\,\,\,$ $v \,=\, -\cos{y}$

We have taken that

$(1).\,\,$ $u = y$

$(2).\,\,$ $dv = \sin{y}\,dy$

We have also evaluated that

$(3).\,\,$ $du \,=\, dy$

$(4).\,\,$ $v \,=\, -\cos{y}$

Now, let’s evaluate $2$ times the indefinite integral of the product of $y$ and $\sin{y}$ with respect to $y$ by the integration by parts formula.

$\implies$ $2 \times \displaystyle \int{y\sin{y}}\,dy$ $\,=\,$ $2$ $\times$ $\bigg(y \times (-\cos{y})$ $-$ $\displaystyle \int{(-\cos{y})}\,dy$ $+$ $k\bigg)$

The literal $k$ denotes the constant of integration on the right hand side of the equation.

$=\,\,$ $2$ $\times$ $\bigg(-y\cos{y}$ $+$ $\displaystyle \int{\cos{y}}\,dy$ $+$ $k\bigg)$

The indefinite integral of cosine of angle $y$ with respect to $y$ is equal to sine of angle $y$ in calculus as per the integral rule of cos function.

$=\,\,$ $2$ $\times$ $(-y\cos{y}$ $+$ $\sin{y}$ $+$ $k)$

The constant of integration $k$ is a constant and the number $2$ is also a constant. So, distribute the number $2$ over the sum and difference of the terms as per distributive property but the keep the trigonometric functions together for expressing it in simple form.

$=\,\,$ $2$ $\times$ $(-y\cos{y}$ $+$ $\sin{y})$ $+$ $2 \times k$

$=\,\,$ $2$ $\times$ $(-y\cos{y}$ $+$ $\sin{y})$ $+$ $2k$

The value of $2k$ is a constant. So, the constant term $2k$ can be simplified denoted by a constant $c$.

$=\,\,$ $2$ $\times$ $(-y\cos{y}$ $+$ $\sin{y})$ $+$ $c$

Now, use the commutative property to change the places of the terms in the second factor of first term in the trigonometric expression.

$=\,\,$ $2$ $\times$ $(\sin{y}$ $-$ $y\cos{y})$ $+$ $c$

The integral of the function is obtained in terms of $y$ by using the integration by parts formula but the given function is defined in terms of $x$. So, it is time to write the integral of function by replacing the variable $y$ by its actual value. Here, we have taken that $y \,=\, \sqrt{x}$

$=\,\,$ $2$ $\times$ $\big(\sin{\sqrt{x}}$ $-$ $\sqrt{x}\cos{\sqrt{x}}\big)$ $+$ $c$

$=\,\,$ $2\big(\sin{\sqrt{x}}$ $-$ $\sqrt{x}\cos{\sqrt{x}}\big)$ $+$ $c$

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