Question

Integrate with respect to x

${e}^{x} ( \frac{1 + sinx}{1 + cosx} )$

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Answer 1

Step-by-step explanation:

Let IIII⟹I=∫ex1−sin(x)1−cos(x)dx=∫ex1−2sin(x2)cos(x2)1−1+2sin2(x2)dx=∫ex1−2sin(x2)cos(x2)2sin2(x2)dx=∫ex(12sin2(x2)−2sin(x2)cos(x2)2sin2(x2))dx=∫ex(csc2(x2)2−cot(x2))dx

Let Jf(x)Jf(x)Jf(x)Jf(x)⟹Jf(x)=∫ex(f(x)+f′(x))dx=∫exf(x)dx+∫exf′(x)dx=f(x)∫exdx−∫f′(x)∫exdxdx+∫exf′(x)dx=exf(x)−∫f′(x)exdx+∫exf′(x)dx+C=exf(x)+C

Let f(x)⟹f′(x)∴J−cot(x2)(x)=−cot(x2)=csc2(x2)2=∫ex(csc2(x2)2−cot(x2))dx

∴I⟹I=J−cot(x2)(x)=−excot(x2)+C

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {e}^{x}\bigg[\dfrac{1 + sinx}{1 + cosx} \bigg] \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm {e}^{x}\bigg[\dfrac{1}{1 + cosx} + \dfrac{sinx}{1 + cosx} \bigg] \: dx$

We know

$\boxed{ \tt{ \: 1 + cos2x = {2cos}^{2}x \: }} \\ \\ \bf \: and \\ \\ \boxed{ \tt{ \: sin2x \: = \: 2sinx \: cosx \: }}$

On substituting these Identities, we get

$\rm \: = \: \displaystyle\int\rm {e}^{x}\bigg[\dfrac{1}{2 {cos}^{2}\dfrac{x}{2}} + \dfrac{2sin\dfrac{x}{2}cos\dfrac{x}{2}}{2 {cos}^{2} \dfrac{x}{2}} \bigg] \: dx$

$\rm \: = \: \displaystyle\int\rm {e}^{x}\bigg[\dfrac{1}{2} {sec}^{2}\dfrac{x}{2} + tan\dfrac{x}{2} \bigg] \: dx$

We know that

$\boxed{ \tt{ \: {e}^{x}(f(x) + f'(x)) \: dx = {e}^{x} \: f(x) \: + \: c \: }} \\ \\ \bf \: and \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}tanx \: = \: {sec}^{2}x \: }}$

So, using these Identities, we get

$\rm \: = \: {e}^{x} \: tan\dfrac{x}{2} \: + \: c \:$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\int\rm {e}^{x}\bigg[\dfrac{1 + sinx}{1 + cosx} \bigg] \: dx = {e}^{x}tan\dfrac{x}{2} + c \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$