Question

integration is this​

Answer 1

Answer:

use sinA-sinB and cosA-cosB

Answer 2

Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{\sin(ax)-\sin(bx)}{\cos(ax)-\cos(bx)}\ dx$

We have,

• $\sin A-\sin B=2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$

• $\cos A-\cos B=-2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$

Then,

$\displaystyle\longrightarrow I=-\int\dfrac{\cos\left(\dfrac{ax+bx}{2}\right)\sin\left(\dfrac{ax-bx}{2}\right)}{\sin\left(\dfrac{ax+bx}{2}\right)\sin\left(\dfrac{ax-bx}{2}\right)}\ dx$

$\displaystyle\longrightarrow I=-\int\dfrac{\cos\left(\left(\dfrac{a+b}{2}\right)x\right)}{\sin\left(\left(\dfrac{a+b}{2}\right)x\right)}\ dx$

$\displaystyle\longrightarrow I=-\dfrac{2}{a+b}\int\dfrac{\left(\dfrac{a+b}{2}\right)\cos\left(\left(\dfrac{a+b}{2}\right)x\right)}{\sin\left(\left(\dfrac{a+b}{2}\right)x\right)}\ dx$

$\displaystyle\longrightarrow\underline{\underline{I=-\dfrac{2}{a+b}\log\left|\sin\left(\dfrac{ax+bx}{2}\right)\right|+C}}$

Hence Integrated!