Question

integrate (sin x + 4)/(sin 2x) dx​

Answer 1

Answer:

Hope this will help you

Step-by-step explanation:

integration of secx and cosec2x isin form of ln

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{sinx + 4}{sin2x} \: dx$

We know,

$\boxed{ \tt{ \: sin2x = 2sinxcosx \: }}$

So, using this identity, we get

$\rm \: = \:\displaystyle\int\rm \frac{sinx + 4}{2sinxcosx} \: dx$

$\rm \: = \:\displaystyle\int\rm \frac{sinx}{2sinxcosx} \: dx \: + \: \displaystyle\int\rm \frac{4}{2sinxcosx} \: dx$

can be rewritten as

$\rm \: = \:\dfrac{1}{2}\displaystyle\int\rm \frac{1}{cosx} \: dx + \displaystyle\int\rm \frac{4}{sin2x} \: dx$

$\rm \: = \:\dfrac{1}{2}\displaystyle\int\rm secx \: dx +4 \displaystyle\int\rm cosec2x \: dx$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm secx \: dx = log |secx + tanx| + c = log \bigg|tan\dfrac{\pi}{4} + \dfrac{x}{2} \bigg| + c}}$

and

$\boxed{ \tt{ \: \displaystyle\int\rm cosecx \: dx = log \bigg|cosecx - cotx\bigg| + c = log \bigg|tan\dfrac{x}{2} \bigg| + c}}$

So, using these Identities, we get

$\rm \: = \:\dfrac{1}{2}log |secx + tanx| + 4 \times \dfrac{1}{2}log |tanx| + c$

$\rm \: = \:\dfrac{1}{2}log |secx + tanx| + 2log |tanx| + c$

OR

$\rm \: = \:\dfrac{1}{2}log \bigg|tan\bigg[\dfrac{\pi}{4} + \dfrac{x}{2} \bigg]\bigg| + 2log |tanx| + c$

More to know :-

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