Question

integrate it .. please no wrong answers.​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Evaluate

$\displaystyle\int\rm {e}^{x} \: \bigg(\dfrac{1 + sinx}{1 + cosx} \bigg) \: dx$

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

Given integral is

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

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

We know,

$\boxed{ \bf{ \: 1 + cos2x = {2cos}^{2}x}}$

and

$\boxed{ \bf{ \: sin2x = 2sinx \: cosx}}$

So, using these, we get

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

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

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

Now, we know,

$\boxed{ \bf{ \: \dfrac{d}{dx}tan\dfrac{x}{2} = \frac{1}{2} {sec}^{2}\dfrac{x}{2}}}$

and

$\boxed{ \bf{ \: \displaystyle\int\rm {e}^{x} \bigg(f(x) + f'(x) \bigg) \: dx \: = \: {e}^{x}f(x) + c}}$

So, using these results, we get

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

Hence,

$\boxed{ \bf{ \: \displaystyle\int\bf {e}^{x} \: \bigg(\dfrac{1 + sinx}{1 + cosx} \bigg) \: dx = {e}^{x} \: tan\dfrac{x}{2} \: + \: c}}$

Additional Information :-

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