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

In this integral problem, the trigonometric functions $\sin{x}$ and $\cos{x}$ formed a rational expression, and we have to evaluate the integration of the trigonometric function with respect to $x$.

$\displaystyle \int{\dfrac{\sin{x}}{\sin{x}+\cos{x}} \,} dx$

Some acceptable adjustments before integration

In this step, we take two mathematically acceptable adjustments, which clear the route for evaluating the integration of the trigonometric function in the upcoming steps.

$= \,\,\,$ $\displaystyle \int{\Bigg(1 \times \dfrac{\sin{x}}{\sin{x}+\cos{x}} \Bigg)\,} dx$

$= \,\,\,$ $\displaystyle \int{\Bigg(\dfrac{2}{2} \times \dfrac{\sin{x}}{\sin{x}+\cos{x}} \Bigg)\,} dx$

$= \,\,\,$ $\displaystyle \int{\Bigg(\dfrac{1 \times 2}{2} \times \dfrac{\sin{x}}{\sin{x}+\cos{x}} \Bigg)\,} dx$

$= \,\,\,$ $\displaystyle \int{\Bigg(\dfrac{1}{2} \times \dfrac{2 \times \sin{x}}{\sin{x}+\cos{x}} \Bigg)\,} dx$

$= \,\,\,$ $\displaystyle \int{\Bigg(\dfrac{1}{2} \times \dfrac{2\sin{x}}{\sin{x}+\cos{x}} \Bigg)\,} dx$

Now, use the constant multiple rule of integration for taking the constant out from the integration.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{ \dfrac{2\sin{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

Previously, we have taken a step for adjusting the function by including the number $2$. Now, we are going to take another step for simplifying this rational expression in trigonometric form.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{ \dfrac{\sin{x}+\sin{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{ \dfrac{\sin{x}+\sin{x}+\cos{x}-\cos{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

Simplify the rational trigonometric expression

Now, we simplify the trigonometric function in rational form and it splits the rational expression into two parts.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{ \dfrac{\sin{x}+\cos{x}+\sin{x}-\cos{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{ \dfrac{(\sin{x}+\cos{x})+(\sin{x}-\cos{x})}{\sin{x}+\cos{x}} \,} dx \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{\Bigg[\dfrac{\sin{x}+\cos{x}}{\sin{x}+\cos{x}} +\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\Bigg] \,} dx \Bigg)$

$= \,\,\,$ $\require{cancel} \dfrac{1}{2} \times \Bigg(\displaystyle \int{\Bigg[\dfrac{\cancel{\sin{x}+\cos{x}}}{\cancel{\sin{x}+\cos{x}}} +\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\Bigg] \,} dx \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{\Bigg[1+\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\Bigg] \,} dx \Bigg)$

Now, use the addition rule of integration to expand the integration to both terms in the expression.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{1 \,}dx+\int{\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

Evaluate the Integration of the functions

Now, we can evaluate the integrations of the both functions mathematically.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \int{}dx+\int{\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1+\displaystyle \int{\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}} \,} dx \Bigg)$

We can notice that the expression in the numerator can be obtained if we differentiate the trigonometric expression in the denominator.

Assume, $u \,=\, \sin{x}+\cos{x}$ and differentiate both sides of the equation with respect to $x$.

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

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\, \sin{x}}+\dfrac{d}{dx}{\, \cos{x}}$

$\implies$ $\dfrac{du}{dx} \,=\, \cos{x}-\sin{x}$

$\implies$ $\dfrac{du}{\cos{x}-\sin{x}} \,=\, dx$

$\implies$ $dx \,=\, \dfrac{du}{\cos{x}-\sin{x}}$

Now, we can substitute the value of differential element $dx$ in the integration for finding the integration of the second term. Then, simplify the expression.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $+$ $\displaystyle \int{\dfrac{\sin{x}-\cos{x}}{u} \,} \times \dfrac{du}{\cos{x}-\sin{x}} \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $+$ $\displaystyle \int{\dfrac{-(\cos{x}-\sin{x})}{u} \,} \times \dfrac{du}{\cos{x}-\sin{x}} \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $+$ $\displaystyle \int{\dfrac{-1 \times (\cos{x}-\sin{x})}{u} \,} \times \dfrac{du}{\cos{x}-\sin{x}} \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $-$ $\displaystyle \int{\dfrac{\cos{x}-\sin{x}}{u} \,} \times \dfrac{du}{\cos{x}-\sin{x}} \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $-$ $\displaystyle \int{\dfrac{\cos{x}-\sin{x}}{u \times (\cos{x}-\sin{x})} \,} du \Bigg)$

$= \,\,\,$ $\require{cancel} \dfrac{1}{2} \times \Bigg(x+c_1$ $-$ $\displaystyle \int{\dfrac{\cancel{\cos{x}-\sin{x}}}{u \times \cancel{(\cos{x}-\sin{x})}} \,} du \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $-$ $\displaystyle \int{\dfrac{1}{u \times 1} \,} du \Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1$ $-$ $\displaystyle \int{\dfrac{1}{u} \,} du \Bigg)$

Now, evaluate the integration of the multiplicative inverse of $u$ by the reciprocal rule of the integration.

$= \,\,\,$ $\dfrac{1}{2} \times \Big(x+c_1-\ln{|u|}+c_2\Big)$

Now, replace the value of variable $u$ to end this problem.

$= \,\,\,$ $\dfrac{1}{2} \times \Big(x+c_1-\ln{|\sin{x}+\cos{x}|}+c_2\Big)$

Now, simplify the expression to obtain the solution of the given integral problem.

$= \,\,\,$ $\dfrac{1}{2} \times \Big(x-\ln{|\sin{x}+\cos{x}|}+c_1+c_2\Big)$

$= \,\,\,$ $\dfrac{1}{2} \times \Big(x-\ln{|\sin{x}+\cos{x}|}+(c_1+c_2)\Big)$

$= \,\,\,$ $\dfrac{1}{2} \times x$ $-$ $\dfrac{1}{2} \times \ln{|\sin{x}+\cos{x}|}$ $+$ $\dfrac{1}{2} \times (c_1+c_2)$

$= \,\,\,$ $\dfrac{1 \times x}{2}$ $-$ $\dfrac{1\times \ln{|\sin{x}+\cos{x}|}}{2}$ $+$ $\dfrac{1\times (c_1+c_2)}{2} $

$= \,\,\,$ $\dfrac{x}{2}$ $-$ $\dfrac{\ln{|\sin{x}+\cos{x}|}}{2}$ $+$ $\dfrac{c_1+c_2}{2} $