Question

If $ \int{\dfrac{\cos x-1}{\sin x+1}{{e}^{x}}dx} $ is equal to
A. $ \dfrac{{{e}^{x}}\cos x}{\sin x+1}+c $
B. $ -\dfrac{{{e}^{x}}\sin x}{\sin x+1}+c $
C. $ \dfrac{-{{e}^{x}}}{\sin x+1}+c $
D. $ -\dfrac{{{e}^{x}}\cos x}{\sin x+1}+c $

Answer 1

Hint: We first discuss the integration by parts method. Integration by parts method is usually used for the multiplication of the functions and their integration. We take two arbitrary functions to express the theorem. We take the $ u=\dfrac{\cos x}{\sin x+1},v={{e}^{x}} $ for our integration $ \int{{{x}^{2}}\ln xdx} $ .

Complete step-by-step answer:
We break the negative part in the numerator of $ \int{\dfrac{\cos x-1}{\sin x+1}{{e}^{x}}dx} $ and get
 $ \int{\dfrac{\cos x-1}{\sin x+1}{{e}^{x}}dx}=\int{\dfrac{{{e}^{x}}\cos x}{\sin x+1}dx}-\int{\dfrac{{{e}^{x}}dx}{\sin x+1}} $ .
We now apply by-parts on the integral form of $ \int{\dfrac{{{e}^{x}}\cos xdx}{\sin x+1}} $ .
Let’s assume $ f\left( x \right)=g\left( x \right)h\left( x \right) $ . We need to find the integration of $ \int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx} $ .
We take $ u=g\left( x \right),v=h\left( x \right) $ . This gives $ \int{f\left( x \right)dx}=\int{uvdx} $ .
The theorem of integration by parts gives $ \int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx} $ .
For our integration $ \int{\dfrac{{{e}^{x}}\cos xdx}{\sin x+1}} $ , we take $ u=\dfrac{\cos x}{\sin x+1},v={{e}^{x}} $ .
Now we complete the integration \[\int{\dfrac{{{e}^{x}}\cos xdx}{\sin x+1}}=\dfrac{\cos x}{\sin x+1}\int{{{e}^{x}}dx}-\int{\left( \dfrac{d\left( \dfrac{\cos x}{\sin x+1} \right)}{dx}\int{{{e}^{x}}dx} \right)dx}\].
The differential part \[\dfrac{d}{dx}\left( \dfrac{\cos x}{\sin x+1} \right)=\dfrac{-\left( \sin x+1 \right)\sin x-{{\cos }^{2}}x}{{{\left( \sin x+1 \right)}^{2}}}=\dfrac{-\sin x-{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\left( \sin x+1 \right)}^{2}}}=\dfrac{-1}{\left( \sin x+1 \right)}\]
We apply these formulas to complete the integration and get
\[\int{\dfrac{{{e}^{x}}\cos xdx}{\sin x+1}}=\dfrac{{{e}^{x}}\cos x}{\sin x+1}+\int{\dfrac{{{e}^{x}}}{\sin x+1}dx}\].
The final form becomes
 $ \begin{align}
  & \int{\dfrac{\cos x-1}{\sin x+1}{{e}^{x}}dx}=\int{\dfrac{{{e}^{x}}\cos x}{\sin x+1}dx}-\int{\dfrac{{{e}^{x}}dx}{\sin x+1}} \\
 & \Rightarrow \int{\dfrac{\cos x-1}{\sin x+1}{{e}^{x}}dx}=\dfrac{{{e}^{x}}\cos x}{\sin x+1}+\int{\dfrac{{{e}^{x}}}{\sin x+1}dx}-\int{\dfrac{{{e}^{x}}dx}{\sin x+1}}=\dfrac{{{e}^{x}}\cos x}{\sin x+1} \\
\end{align} $
The correct option is A.
So, the correct answer is “Option A”.

Note: In case one of two functions are missing and we need to form the by parts method, we will take the multiplying constant 1 as the second function.
For example: if we need to find \[\int{\ln xdx}\], we have only one function. So, we take constant 1 as the second function where $ u=\ln x,v=1 $ . But we need to remember that we won’t perform by parts by taking $ u=1 $ .