Question

integrate (2sin x)/(1 + cos 2x) dx​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{2sinx}{1 + cos2x} \: dx$

We know,

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

So, using this identity, the above integral can be rewritten as

$\rm \: = \:\displaystyle\int\rm \frac{2sinx}{ {2cos}^{2} x} \: dx$

$\rm \: = \:\displaystyle\int\rm \frac{sinx}{cosx \times cosx} \: dx$

$\rm \: = \:\displaystyle\int\rm \bigg[\dfrac{1}{cosx} \times \dfrac{sinx}{cosx} \bigg] \: dx$

$\rm \: = \:\displaystyle\int\rm [ \: secx \: tanx \: ] \: dx$

$\rm \: = \:secx \: + \: c$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\int\rm \frac{2sinx}{1 + cos2x} \: dx \: = \: secx \: + \: c \: }}$

More to know :-

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

Answer 2

Answer:

$</p><p>\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\int\rm \frac{2sinx}{1 + cos2x} \: dx \: = \: secx \: + \: c \: }}</p><p>$

Step-by-step explanation:

Hope it helps you