Evaluate the following integrals:

sin(A - B) = sinAcosB - cosAsinB

∴ We can write

sin(A + B) = sinAcosB + cosAsinB

∴ We can write

∴ The given equation becomes

⇒

⇒

Denominator is of the form (a - b)(a + b) = a² - b²

⇒ ….(1)

We know sin²x + cos²x = 1

∴ sin²x =1 - cos²x

Substituting the above result in (1) we get

⇒

⇒ …(2)

Let us assume

⇒

⇒ 2sinx.cosx.dx=dt

⇒ sin2x.dx=dt

Substituting dt and t in (2) we get

⇒

= ln|t| + c

But t =

∴ ln| | + c.