Question

Suppose $J = \int {\dfrac{{{{\sin }^2}x + \sin x}}{{1 + \sin x + \cos x}}} dx$ and $K = \int {\dfrac{{{{\cos }^2}x + \cos x}}{{1 + \sin x + \cos x}}} dx$. If $C$ is an arbitrary constant of integration, then which of the following is/are correct?
(a) $J = \dfrac{1}{2}\left( {x - \sin x + \cos x} \right) + C$
(b) $J = K - \left( {\sin x + \cos x} \right) + C$
(c) $J = x - K + C$
(d) $K = \dfrac{1}{2}\left( {x - \sin x + \cos x} \right) + C$

Answer 1

Here, we need to find which of the given options is true. We will add the given equations and use the sum of integrals. Then, we will simplify and integrate the resulting expression. Finally, we will rewrite the expression to get the required answer.
Formula Used: The square of the sine and cosine of an angle $\theta$ is always equal to 1, that is ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

Complete step by step solution:
We will add the given equations and simplify to get the required answer.
Adding the given equations $J = \int {\dfrac{{{{\sin }^2}x + \sin x}}{{1 + \sin x + \cos x}}} dx$ and $K = \int {\dfrac{{{{\cos }^2}x + \cos x}}{{1 + \sin x + \cos x}}} dx$, we get
$\Rightarrow J + K = \int {\dfrac{{{{\sin }^2}x + \sin x}}{{1 + \sin x + \cos x}}} dx + \int {\dfrac{{{{\cos }^2}x + \cos x}}{{1 + \sin x + \cos x}}} dx$
Therefore, adding the two integrals, we get
$\Rightarrow J + K = \int {\left( {\dfrac{{{{\sin }^2}x + \sin x}}{{1 + \sin x + \cos x}} + \dfrac{{{{\cos }^2}x + \cos x}}{{1 + \sin x + \cos x}}} \right)} dx$
The denominators of both the functions in the parentheses is the same.
Adding the two functions, we get
$\Rightarrow J + K = \int {\left( {\dfrac{{{{\sin }^2}x + \sin x + {{\cos }^2}x + \cos x}}{{1 + \sin x + \cos x}}} \right)} dx$
Rewriting the equation, we get
$\Rightarrow J + K = \int {\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x + \sin x + \cos x}}{{1 + \sin x + \cos x}}} \right)} dx$
We know that the square of the sine and cosine of an angle $\theta$ is always equal to 1, that is ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Therefore, we get
$\Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
Substituting ${\sin ^2}x + {\cos ^2}x = 1$ in the equation $J + K = \int {\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x + \sin x + \cos x}}{{1 + \sin x + \cos x}}} \right)} dx$, we get
$\Rightarrow J + K = \int {\left( {\dfrac{{1 + \sin x + \cos x}}{{1 + \sin x + \cos x}}} \right)} dx$
The numerator and the denominator are equal. Therefore, we can simplify the equation as
$\Rightarrow J + K = \int {\left( 1 \right)} dx$
Now, we will integrate the expression.
The integral of a constant $\int {\left( 1 \right)} dx$ can be written as $\int {\left( 1 \right)} dx = x + C$, where $C$ is an arbitrary constant of integration.
Therefore, the equation becomes
$\Rightarrow J + K = x + C$
Subtracting $K$ from both sides of the equation, we get
$\Rightarrow J + K - K = x + C - K$
Thus, we get
$\Rightarrow J = x - K + C$
$\therefore$ The equation $J = x - K + C$ is correct, where $C$ is an arbitrary constant of integration.

The correct option is option (c).

Note:
We used the sum of two integrals to add the integrals $\int {\dfrac{{{{\sin }^2}x + \sin x}}{{1 + \sin x + \cos x}}} dx$ and $\int {\dfrac{{{{\cos }^2}x + \cos x}}{{1 + \sin x + \cos x}}} dx$. The sum of the integrals of two functions $f\left( x \right)$ and $g\left( x \right)$ is equal to the integral of the sum of the two functions, that is $\int {f\left( x \right)} dx + \int {g\left( x \right)} dx = \int {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx$.