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 $x$.

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

Firstly, let’s concentrate on differentiating the expression on the left-hand side of the equation. So, let’s differentiate the $u$ square with respect to $x$.

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

In this problem, $u$ and $x$ both are variables. So, the derivative of $u$ square with respect to $x$ should not be calculated by considering $u$ as a constant, whereas the derivative of $u^2$ can be evaluated with respect to $x$ by the chain rule.

$\implies$ $\dfrac{d}{dx} \times 1 \times {\,u^2}$ $\,=\,$ $\dfrac{d}{dx}{(3+2\cos{x})}$

The function $u$ square is defined in terms of $u$ and it should be differentiated with respect to $u$. So, let’s try to include the differential element in terms of $u$.

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

Now, use the multiplication of the fractions and then commutative property to change their numerators and it helps us to find the differentiation on the left-hand side of the equation.

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

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

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

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

Now, use the power rule of differentiation to find the derivative of $u$ square with respect to $u$.

$\implies$ $\dfrac{du}{dx} \times 2u^{2-1}$ $\,=\,$ $\dfrac{d}{dx}{(3+2\cos{x})}$

$\implies$ $\dfrac{du}{dx} \times 2u^1$ $\,=\,$ $\dfrac{d}{dx}{(3+2\cos{x})}$

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

Now, use the commutative property one more time to change the positions of the factors on the left-hand side of the equation.

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

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

Now, it is time to find the differentiation on the right-hand side of the equation. Look at the function on the right-hand side of the equation, and it expresses sum of two functions. So, left’s find the derivative of sum of two functions by the addition rule of derivatives.

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

According to derivative of constant rule, the derivative of three with respect to $x$ is equal to zero.

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

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

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

The number $2$ is a constant and it can be taken out from the differentiation. Use the constant multiple derivative rule to separate the number $2$ from the differentiation.

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

As per the derivative of sine function rule, the derivative of cosine of angle $x$ with respect to $x$ is equal to negative sine of angle $x$.

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

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

Look at the expressions on both sides of the equation. The number $2$ is a coefficient and it can be eliminated from both expressions by dividing them with number $2$.

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

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

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

It is time to simplify the whole equation to convert the function in terms of $x$ into the function in terms of $u$.

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

Let’s express each factor on both sides in fraction form and then multiply the fractions to find their product as per the multiplication of fractions.

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

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

$\implies$ $\dfrac{udu}{dx}$ $\,=\,$ $\dfrac{(-\sin{x})}{1}$

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

$\implies$ $udu \times 1$ $\,=\,$ $(-\sin{x}) \times dx$

$\implies$ $udu$ $\,=\,$ $(-1 \times \sin{x}) \times dx$

$\implies$ $udu$ $\,=\,$ $(-1) \times \sin{x}dx$

$\implies$ $(-1) \times \sin{x}dx$ $\,=\,$ $udu$

Now, divide the expressions on both sides of the equation by the integer negative one to eliminate the $-1$ from the expression on the left-hand side of the equation.

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

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

$\,\,\,\,\,\,\therefore\,\,\,$ $\sin{x}dx$ $\,=\,$ $\dfrac{udu}{(-1)}$

We have taken that $u^2 \,=\, 3+2\cos{x}$ and we have already derived that $\sin{x}dx$ $\,=\,$ $\dfrac{udu}{(-1)}$.

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

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

It is time to simplify the function in terms of $u$.

$\,=\,\,$ $\displaystyle \int{\dfrac{1}{u}} \times \,\dfrac{udu}{(-1)}$

Let’s use the multiplication of fractions rule to split a fraction into two fractions.

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

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

Now, multiply the first two fractions to find their product and it helps us to simplify the function.

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

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

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

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

The number $1$ can be written as a product of two negative ones and it helps us to remove $-1$ from denominator.

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

$\,=\,\,$ $\displaystyle \int{\dfrac{\cancel{(-1)} \times (-1)}{\cancel{(-1)}}} \times du$

$\,=\,\,$ $\displaystyle \int{(-1)} \times du$

The integer $-1$ is a constant and it does not play any role while finding the integration. So, the integer can be excluded from the integration as per the constant multiple integral rule.

$\,=\,\,$ $(-1) \times \displaystyle \int{}\,du$

The number is a factor of every quantity. So, number one can be written as a factor of the differential element $du$.

$\,=\,\,$ $(-1) \times \displaystyle \int{1} \times \,du$

$\,=\,\,$ $(-1) \times \displaystyle \int{1}\,du$

Now, the integral of $1$ with respect to $u$ can be evaluated by the integral of one formula. Here, $k$ is the constant of integration

$\,=\,\,$ $(-1) \times (u+k)$

Use the distributive property to distribute the $-1$ to both terms.

$\,=\,\,$ $(-1) \times u$ $+$ $(-1) \times k$

$\,=\,\,$ $-u$ $+$ $(-1) \times k$

In this problem, $k$ is a constant and $-1$ is also a constant. So, their product can be simply denoted by a constant $c$.

$\,=\,\,$ $-u+c$

The integration is successfully completed and it is time to convert the function in terms of $u$ into the function in terms of $x$.

We know that $u^2$ $\,=\,$ $3+2\cos{x}$. Therefore, $u$ $\,=\,$ $\sqrt{3+2\cos{x}}$. Now, substitute the value of $u$ in expression to get the integral of the given irrational function.

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