# Evaluate $\displaystyle \int{\sin{\sqrt{x}}}\,dx$

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 $x$.

### Reducing the complexity by the change of variable

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 $y$.

### Find integral of function with integration by parts

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.

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

### Express the integration of function in actual form

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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