Integral Calculus
86267
\(\int \frac{\cos x-1}{\sin x+1} e^{x} d x\) is equal to:
1 \(\frac{e^{x} \cos \mathrm{x}}{1+\sin \mathrm{x}}+\mathrm{c}\)
2 \(c-\frac{e^{x} \sin x}{1+\sin x}\)
3 \(c-\frac{e^{x}}{1+\sin \mathrm{x}}\)
4 \(c-\frac{e^{x} \cos x}{1+\sin x}\)
Explanation:
(A) : Let,
\(I=\int \frac{\cos x-1}{\sin x+1} e^{x} \cdot d x\)
\(=\int e^{x}\left(\frac{\cos x}{\sin x+1}-\frac{1}{\sin x+1}\right) d x\)
\(\left[\because \int \mathrm{e}^{\mathrm{x}}\left[\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right] \mathrm{dx}=\mathrm{e}^{\mathrm{x}} . \mathrm{f}(\mathrm{x})+\mathrm{C}\right.\)
\(\therefore \quad \mathrm{I}=\mathrm{e}^{\mathrm{x}} \frac{\cos \mathrm{x}}{\sin \mathrm{x}+1}+\mathrm{C}\)