Question

integration sin x / sin (x-a) dx​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\int \sf \: \dfrac{sinx}{sin(x - a)} \: dx$

☆ On adding and Subtracting 'a' in angle of numerator,

$\rm \:= \: \:\:\displaystyle\int \sf \: \dfrac{sin(x - a + a)}{sin(x - a)} \: dx$

$\rm \:= \: \:\:\displaystyle\int \sf \: \dfrac{sin \bigg((x - a )+ a \bigg)}{sin(x - a)} \: dx$

$\rm \:= \: \:\:\displaystyle\int \sf \: \dfrac{sin(x - a)cosa + sina \: cos(x - a)}{sin(x - a)} \: dx$

$\red{\bigg \{ \because \: sin(x + y) = sinxcosy + sinycosx\bigg \}}$

$\rm \:= \: \:\displaystyle\int \sf \: \dfrac{ \cancel{sin(x - a)} \: cosa}{ \cancel{sin(x - a)}} dx + \displaystyle\int \sf \: \dfrac{sina \: cos(x - a)}{sin(x - a)} dx$

$\rm \:= \: \:\displaystyle\int \sf \: cosa \: dx \: + \: \displaystyle\int \sf \: sina \: cot(x - a) \: dx$

$\red{\bigg \{ \because \:\dfrac{cosx}{sinx} = cotx \bigg \}}$

$\rm \:= \: \:(cosa) \: x +sina \: logsin(x - a) + c$

$\red{\bigg \{ \because \: \displaystyle\int \sf \: kdx \: = \: x \: + \: c\bigg \}} \\ \red{\bigg \{ \because \: \displaystyle\int \sf \: cotx = log(sinx) + c\bigg \}}$

Additional Information :-

$\green{\boxed{\bf{\displaystyle\int \sf \: sinx = - cosx + c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: cosx = sinx + c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: tanx = log \: secx + c = - log \: cosx+ c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: cotx = log \: sinx + c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: secxtanx = secx+ c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: cosecxcotx = - cosecx+ c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: {sec}^{2}x = tanx+ c }}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: {cosec}^{2}x = - cotx+ c }}}$