Integrate sinx/sinx-cosx..... Plz note sinx-cosx as a whole is in the denominator..
use the good old quotient rule:
if y = u/v, y' = (u'v - uv')/v^2
u = sinx
v = sinx - cosx
y' = [(cosx)(sinx-cosx) - (sinx)(cosx+sinx)]/(sinx-cosx)^2
you can play with that some, but I like
1/(sin(2x)-1)
To integrate the given expression, we can use a trigonometric substitution method. Let's start by simplifying the expression in the denominator.
The denominator, sinx - cosx, can be rewritten using the trigonometric identity for the difference of angles:
sinx - cosx = √2(sin(x - π/4))
Now, let's substitute u = x - π/4:
This implies that du = dx, as the derivative of u with respect to x is 1.
Now, we can rewrite the denominator as √2 sin(u):
√2 sin(u)
The integral can now be expressed as:
∫ (sinx / (sinx - cosx)) dx = ∫ (sinx / (√2 sin(u))) dx
To simplify further, we can cancel out sin(x) from the numerator and denominator:
∫ (1 / √2) du
Now, we can integrate the simplified expression:
∫ (1 / √2) du = (1 / √2) ∫ du
Integrating ∫ du yields:
(1 / √2) u + C
Substituting u back with x - π/4, we get the final answer:
(1 / √2) (x - π/4) + C
Therefore, the integral of sinx / (sinx - cosx) with respect to x is:
(1 / √2) (x - π/4) + C, where C is the constant of integration.