Question

integrate (sin(x - a))/(sin(x - b)) dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\int\dfrac{sin(x-a)}{sin(x-b)}\,dx}$

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

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

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

$\tt{\leadsto\,We\,\,know,\,\,\green{sin(\alpha+\beta)=sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)}}$

So,

$\sf{=\displaystyle\int\dfrac{sin(x-b)cos(b-a)+cos(x-b)sin(b-a)}{sin(x-b)}\,dx}$

$\sf{=\displaystyle\int\bigg\{\dfrac{sin(x-b)cos(b-a)}{sin(x-b)}+\dfrac{cos(x-b)sin(b-a)}{sin(x-b)}\bigg\}\,dx}$

$\sf{=\displaystyle\int\bigg\{cos(b-a)+cot(x-b)sin(b-a)\bigg\}\,dx}$

$\sf{=\displaystyle\int\,cos(b-a)\,dx+\int\,cot(x-b)sin(b-a)\,dx}$

$\sf{=\displaystyle\,cos(b-a)\int\,dx+sin(b-a)\int\,cot(x-b)\,dx}$

$\tt{\leadsto\,We\,\,know,\,\,\green{\displaystyle\,\int\,dx=x+c\,\,\,\,\,and\,\,\,\,\,\int\,cot(x)\,dx=\ln|sin(x)|+c}}$

So,

$\sf{=\displaystyle\,cos(b-a)\cdot\,x+sin(b-a)\,\cdot\ln|sin(x-b)|+C}$

$\sf{=\displaystyle\,x\,cos(b-a)+sin(b-a)\,\ln|sin(x-b)|+C}$