∫cos⁡x−1sin⁡x+1exdx is equal to - Infinity Learn by Sri Chaitanya

A
$\frac{e^{x} \cos x}{1 + \sin x} + C$
B
$C - \frac{e^{x} \sin x}{1 + \sin x}$
C
$C - \frac{e^{x}}{1 + \sin x}$
D
$C - \frac{e^{x} \cos x}{1 + \sin x}$

Solution:

$\begin{array}{l} \int \frac{\cos x - 1}{\sin x + 1} e^{x} dx \\ = \int \frac{\cos x}{1 + \sin x} e^{x} dx - \int \frac{1}{1 + \sin x} e^{x} dx \\ = \frac{(\cos x) e^{x}}{1 + \sin x} - \int \frac{- (1 + \sin x) \sin x - \cos^{2} x}{(1 + \sin x)^{2}} e^{x} dx - \int \frac{e^{x}}{\sin x + 1} dx + C \\ = \frac{e^{x} \cos x}{1 + \sin x} + \int \frac{1}{1 + \sin x} e^{x} dx - \int \frac{e^{x}}{1 + \sin x} dx + C \\ = \frac{e^{x} \cos x}{1 + \sin x} + C \end{array}$

$\int \frac{\cos x - 1}{\sin x + 1} e^{x} dx$ is equal to

Solution:

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