Question

Evaluate the given indefinite integral: $\int{\dfrac{\sin x}{\sin x-\cos x}dx}$?
(a) $\dfrac{1}{2}\log \left( \sin x-\cos x \right)+x+c$
(b) $\dfrac{1}{2}\left( \log \left( \sin x-\cos x \right)+x \right)+c$
(c) $\dfrac{1}{2}\log \left( \cos x-\sin x \right)+x+c$
(d) $\dfrac{1}{2}\left( \log \left( \cos x-\sin x \right)+x \right)+c$

Answer 1

We start solving the problem by assigning the variable for the given indefinite integral. We then make the necessary arrangements to make it as a sum of two integrals. We then solve each integral separately and then add them together at last. We solve the first integral by using the property that $\int{adx}=ax+C$, where C is an arbitrary constant. We then solve the second integral by using the facts that $d\left( \sin x \right)=\cos xdx$, $d\left( \cos x \right)=-\sin xdx$ and $\int{\dfrac{1}{x}dx}=\log x+C$, where C is an arbitrary constant. We then add both the obtained integrals to get the required result.

Complete step-by-step answer:
According to the problem, we are asked to evaluate the given indefinite integral $\int{\dfrac{\sin x}{\sin x-\cos x}dx}$.
Let us assume $I=\int{\dfrac{\sin x}{\sin x-\cos x}dx}$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x-\cos x}dx}$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x-\cos x+\cos x}{\sin x-\cos x}dx}$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\sin x-\cos x+\sin x+\cos x}{\sin x-\cos x}dx}$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\sin x-\cos x}{\sin x-\cos x}dx}+\dfrac{1}{2}\int{\dfrac{\sin x+\cos x}{\sin x-\cos x}dx}$.
Let us assume $\int{\dfrac{\sin x-\cos x}{\sin x-\cos x}dx}={{I}_{1}}$ and $\int{\dfrac{\sin x+\cos x}{\sin x-\cos x}dx}={{I}_{2}}$.
$\Rightarrow I=\dfrac{1}{2}{{I}_{1}}+\dfrac{1}{2}{{I}_{2}}$ ---(1).
Let us first solve ${{I}_{1}}$, ${{I}_{2}}$ and then substitute the results in equation (1).
Now, let us solve for ${{I}_{1}}$.
So, we have ${{I}_{1}}=\int{\dfrac{\sin x-\cos x}{\sin x-\cos x}dx}$.
$\Rightarrow {{I}_{1}}=\int{1dx}$.
We know that $\int{adx}=ax+C$, where C is an arbitrary constant.
$\Rightarrow {{I}_{1}}=x+{{c}_{1}}$ ---(2).
Now, let us solve for ${{I}_{2}}$.
So, we have ${{I}_{2}}=\int{\dfrac{\sin x+\cos x}{\sin x-\cos x}dx}$ ---(3).
Let us assume $\sin x-\cos x=t$ ---(4). Let us apply differential on both sides.
So, we get $d\left( \sin x-\cos x \right)=dt$.
We know that $d\left( a-b \right)=da-db$.
$\Rightarrow d\left( \sin x \right)-d\left( \cos x \right)=dt$.
We know that $d\left( \sin x \right)=\cos xdx$ and $d\left( \cos x \right)=-\sin xdx$.
$\Rightarrow \left( \cos xdx \right)-\left( -\sin xdx \right)=dt$.
$\Rightarrow \left( \cos x+\sin x \right)dx=dt$ ---(5).
Let us substitute equations (4) and (5) in equation (3).
So, we get ${{I}_{2}}=\int{\dfrac{1}{t}dt}$.
We know that $\int{\dfrac{1}{x}dx}=\log x+C$, where C is an arbitrary constant.
$\Rightarrow {{I}_{2}}=\log t+{{c}_{2}}$.
From equation (4), we have $t=\sin x-\cos x$.
$\Rightarrow {{I}_{2}}=\log \left( \sin x-\cos x \right)+{{c}_{2}}$ ---(6).
Let us substitute equation (2) and (6) in equation (1).
So, we get $I=\dfrac{1}{2}\left( x+{{c}_{1}} \right)+\dfrac{1}{2}\left( \log \left( \sin x-\cos x \right)+{{c}_{2}} \right)$.
$\Rightarrow I=\dfrac{1}{2}\left( \log \left( \sin x-\cos x \right)+x \right)+\dfrac{1}{2}\left( {{c}_{1}}+{{c}_{2}} \right)$.
Since ${{c}_{1}}$, ${{c}_{2}}$ are two arbitrary constants, we replace $\dfrac{{{c}_{1}}+{{c}_{2}}}{2}$ with $c$.
$\Rightarrow I=\dfrac{1}{2}\left( \log \left( \sin x-\cos x \right)+x \right)+c$.
So, we have found the result of indefinite integral $\int{\dfrac{\sin x}{\sin x-\cos x}dx}$ as $\dfrac{1}{2}\left( \log \left( \sin x-\cos x \right)+x \right)+c$.

So, the correct answer is “Option B”.

Note: We can see that the given problems contain a huge amount of calculation, so we need to perform each step to avoid calculation mistakes. We should not forget to add arbitrary constants while solving problems that consist of indefinite integrals. We can also make use of the formula $\int{\dfrac{d\sin x+e\cos x}{f\sin x+g\cos x}dx}=a\log \left( f\sin x+g\sin x \right)+bx+C$, where $a\left( \dfrac{d}{dx}\left( f\sin x+g\cos x \right) \right)+b\left( f\sin x+g\cos x \right)=d\sin x+e\cos x$. We can make use of the results $\cos x=\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}$ and $\sin x=\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}$ to get the required result.