Evaluate $\displaystyle \int{\dfrac{\sin{x}}{\sqrt{3+2\cos{x}}}}\,dx$

Now, let’s differentiate the expressions on both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(u)}$ $\,=\,$ $\dfrac{d}{dx}{(3+2\cos{x})}$

The letter $u$ is a variable and it is not $x$ but it represents a function in terms of $x$. So, the derivative of $u$ with respect to $x$ cannot be evaluated. So, it is simply written as derivative of $u$ with respect to $x$ on the left-hand side of the expression.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(3+2\cos{x})}$

Now, let’s focus on finding the differentiation of the expression on the right-hand side of the equation. The expression is formed by the addition of two functions. So, the derivative of sum of two functions can be evaluated by the addition rule of derivatives.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(3)}$ $+$ $\dfrac{d}{dx}{(2\cos{x})}$

The number $3$ is a constant and its derivative is zero as per the derivative of a constant formula.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $0$ $+$ $\dfrac{d}{dx}{(2\cos{x})}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(2\cos{x})}$

The number $2$ is a coefficient of cosine function and it is also a constant. Actually, it does not interrupt the integration. So, it can be separated from the differentiation.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(2 \times \cos{x})}$

Now, let’s separate the constant $2$ from the integration by the constant multiple rule of integrals.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2 \times \dfrac{d}{dx}{\cos{x}}$

Use, the derivative of sine function formula to find the derivative of sine of angle $x$ with respect to $x$.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2 \times (-\sin{x})$

Let’s prepare the equation for cross multiplication and it helps us to find the differential element $dx$.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $2 \times (-1) \times \sin{x}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $(-2) \times \sin{x}$

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $\dfrac{(-2) \times \sin{x}}{1}$

Now, use cross multiply rule to simplify the equation further.

$\implies$ $du \times 1$ $\,=\,$ $\big((-2) \times \sin{x}\big) \times dx$

$\implies$ $du$ $\,=\,$ $(-2) \times (\sin{x} \times dx)$

The equation can be written as follows.

$\implies$ $(-2) \times (\sin{x} \times dx)$ $\,=\,$ $du$

The numerical factor $-2$ is multiplied at left-hand side of the equation and it will divide the expression on the right-hand side of the equation.

$\implies$ $\sin{x} \times dx$ $\,=\,$ $\dfrac{du}{(-2)}$

Leave the sine function on the left hand side of the equation because sine function and differential element dx get multiplied in the given function.

$\,\,\,\,\,\,\therefore\,\,\,$ $\sin{x}dx$ $\,=\,$ $\dfrac{du}{(-2)}$

We have taken that $u \,=\, 3+2\cos{x}$ and it is derived that $\sin{x}dx$ $\,=\,$ $\dfrac{du}{(-2)}$.

Now, substitute them in the above function to convert the expression in terms of $x$ into an expression in terms of $u$.

$\implies$ $\displaystyle \int{\dfrac{1}{\sqrt{3+2\cos{x}}} \times \sin{x}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}}} \times \,\dfrac{du}{(-2)}$

It is time to find the integral of the function with respect to $u$.

$\,=\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}}} \times \,\dfrac{du}{(-2)}$

$\,=\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}}} \times \,\dfrac{1 \times du}{(-2) \times 1}$

$\,=\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}}} \times \,\dfrac{1}{(-2)} \times \,\dfrac{du}{1}$

$\,=\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}}} \times \,\dfrac{1}{(-2)} \times \,du$

$\,=\,\,$ $\displaystyle \int{\dfrac{1}{(-2)} \times \dfrac{1}{\sqrt{u}}} \times \,du$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \displaystyle \int{\dfrac{1}{\sqrt{u}}} \times \,du$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \displaystyle \int{\dfrac{1}{\sqrt{u}}\,}du$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \displaystyle \int{\dfrac{1}{u^{\Large \frac{1}{2}}}\,}du$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \displaystyle \int{u^{- \Large \frac{1}{2}}\,}du$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \Bigg(\dfrac{u^{- {\Large \frac{1}{2}}+1}}{-\dfrac{1}{2}+1}+c_1\Bigg)$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \Bigg(\dfrac{u^{\Large \frac{-1+2}{2}}}{\dfrac{-1+2}{2}}+c_1\Bigg)$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \Bigg(\dfrac{u^{\Large \frac{1}{2}}}{\dfrac{1}{2}}+c_1\Bigg)$

$\,=\,\,$ $\dfrac{1}{(-2)} \times \dfrac{u^{\Large \frac{1}{2}}}{\dfrac{1}{2}}$ $+$ $\dfrac{1}{(-2)} \times c_1$

$\,=\,\,$ $\dfrac{1 \times u^{\Large \frac{1}{2}}}{(-2) \times \dfrac{1}{2}}$ $+$ $\dfrac{1}{(-2)} \times c_1$

$\,=\,\,$ $\dfrac{1 \times u^{\Large \frac{1}{2}}}{(-2) \times \dfrac{1}{2}}$ $+$ $\dfrac{1}{(-2)} \times c_1$

$\,=\,\,$ $\dfrac{u^{\Large \frac{1}{2}}}{(-2) \times \dfrac{1}{2}}$ $+$ $\dfrac{1}{(-2)} \times c_1$

$\,=\,\,$ $\dfrac{u^{\Large \frac{1}{2}}}{\dfrac{(-2)}{1} \times \dfrac{1}{2}}$ $+$ $\dfrac{1}{(-2)} \times \dfrac{c_1}{1}$

$\,=\,\,$ $\dfrac{u^{\Large \frac{1}{2}}}{\dfrac{(-2) \times 1}{1 \times 2}}$ $+$ $\dfrac{1 \times c_1}{(-2) \times 1}$

$\,=\,\,$ $\dfrac{u^{\Large \frac{1}{2}}}{\dfrac{(-2)}{2}}$ $+$ $\dfrac{c_1}{(-2)}$

$\,=\,\,$ $\dfrac{u^{\Large \frac{1}{2}}}{\dfrac{\cancel{(-2)}}{\cancel{2}}}$ $+$ $\dfrac{c_1}{(-2)}$

$\,=\,\,$ $\dfrac{u^{\Large \frac{1}{2}}}{-1}$ $+$ $\dfrac{c_1}{(-2)}$

$\,=\,\,$ $-u^{\Large \frac{1}{2}}$ $+$ $\dfrac{c_1}{(-2)}$

$\,=\,\,$ $-u^{\Large \frac{1}{2}}+c$

$\,=\,\,$ $-\sqrt{u}+c$

$\,=\,\,$ $-\sqrt{3+2\cos{x}}+c$