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

\frac{e^{x} cos x}{1 + sin x} + c

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c - \frac{e^{x} sin x}{1 + sin x}

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c - \frac{e^{x}}{1 + sin x}

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c - \frac{e^{x} c o s x}{1 + sin x}

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Solution

The correct option is A $\frac{e^{x} cos x}{1 + sin x} + c$
Let $u$ be $\frac{c o s x - 1}{s i n x + 1}$
and $d v$ be $e^{x}$ then $v = e^{x}$
we know that $\int u d v = u v - \int v d u$
$\Rightarrow \int (\frac{c o s x - 1}{s i n x + 1}) e^{x}$ = $(\frac{c o s x - 1}{s i n x + 1}) e^{x} - \int e^{x} \frac{(c o s x - 1) (c o s x) - (s i n x + 1) (- s i n x)}{(s i n x + 1)^{2}}$ \
we have to do the by parts again taking $u = \frac{(c o s x - 1) (c o s x) - (s i n x + 1) (- s i n x)}{(s i n x + 1)^{2}}$ and $v = e^{x}$
$\Rightarrow \int (\frac{1 - c o s x + s i n x}{(1 + s i n x)^{2}}) e^{x} = (\frac{1 - c o s x + s i n x}{(1 + s i n x)^{2}}) e^{x} - \int e^{x} (\frac{(1 - c o s x + s i n x) 2 (1 + s i n x) (c o s x) - (1 + s i n x)^{2} (s i n x + c o s x)}{(1 + s i n x)^{4}})$
$\Rightarrow \int (\frac{1 - c o s x + s i n x}{(1 + s i n x)^{2}}) e^{x} = - (\frac{e^{x}}{1 + s i n x})$ on back substitution we get
$\Rightarrow \int (\frac{c o s x - 1}{s i n x + 1}) e^{x}$ = $(\frac{c o s x - 1}{s i n x + 1}) e^{x} - (- (\frac{e^{x}}{1 + s i n x}))$
$\Rightarrow \int (\frac{c o s x - 1}{s i n x + 1}) e^{x}$ = $\frac{e^{x} (c o s x)}{1 + s i n x}$
Hence, option, 'A' is correct.

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