Question

Evaluate the integral and then fill in the blanks for $\int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=.....+C$
A. $-\log \left| x\sin x-\cos x-1 \right|$
B. $\log \left| x\sin x-\cos x-1 \right|$
C. $-\log \left| x\cos x-\sin x-1 \right|$
D. $\log \left| x\cos x-\sin x-1 \right|$

Answer 1

In this problem we have to calculate the integral value of the given equation. From this we are going to use the substitution method by substituting $u=x\cos x-\sin x-1$ and calculate the value of $du$. We will use the $uv$ formula of differentiation which is ${{\left( uv \right)}^{'}}=u{{v}^{'}}+v{{u}^{'}}$ and simplify the equation to get the value of $du$. After getting the value of $du$, we will substitute $u$, $du$ in the given integration and simplify the equation. Now we will use the integration formula $\int{\dfrac{dx}{x}}=\log \left| x \right|+C$ and calculate the required value.

Complete step by step solution:
Given that, $\int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=.....+C$.
Considering the integration part which is $\int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}$.
To solve the above integration part, we are going to use the substitution method. So, we are going to substitute $u=x\cos x-\sin x-1$ in the integration part. Before substituting this value, we also need the value of $du$. For this we are going to differentiate the value $u=x\cos x-\sin x-1$ with respect to $x$, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( x\cos x-\sin x-1 \right)$
Applying the differentiation for each term, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( x\cos x \right)-\dfrac{d}{dx}\left( \sin x \right)-\dfrac{d}{dx}\left( 1 \right)$
We have the differentiation formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\dfrac{d}{dx}\left( 1 \right)=0$. Substituting these values in the above equation, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( x\cos x \right)-\cos x$
Using the differentiation formula ${{\left( uv \right)}^{'}}=u{{v}^{'}}+v{{u}^{'}}$ in the above equation, then we will get
$\Rightarrow \dfrac{du}{dx}=x\dfrac{d}{dx}\left( \cos x \right)+\cos x\dfrac{d}{dx}\left( x \right)-\cos x$
We know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, $\dfrac{d}{dx}\left( x \right)=1$. Substituting these values and simplifying the equation, then we will get
$\begin{aligned} & \Rightarrow \dfrac{du}{dx}=x\left( -\sin x \right)+\cos x-\cos x \\\ & \Rightarrow du=-x\sin xdx \\\ \end{aligned}$
From the values $u=x\cos x-\sin x-1$, $du=-x\sin xdx$, the given integration value modified as
$\begin{aligned} & \Rightarrow \int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=\int{\dfrac{1}{u}\left( -du \right)} \\\ & \Rightarrow \int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=-\int{\dfrac{du}{u}} \\\ \end{aligned}$
We know that $\int{\dfrac{dx}{x}=\log \left| x \right|+C}$, then we will have
$\Rightarrow \int{\dfrac{x\sin x}{x\cos x-\sin x-1}dx}=-\log \left| x\cos x-\sin x-1 \right|+C$
Hence option – C is the correct answer.

Note: We can also use the formula $\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx}=\log \left| f\left( x \right) \right|+C$ for the above problem. We can use this formula after calculating the derivative of the denominator and use this formula to get the required result.