Evaluate the integral of (sin 2x)/(1+cos^2 x)
1. u=cosx and du=-sinx *dx
2. evaluate the integral of -1/(1+u^2)*du
3. result is -ln(1+u^2)+C
Where did the sin2x dissapear too???
recall that sin2x = 2 sinx cosx
(sin2x)/(1+cos^2x) dx
= 2cosx/(1+cos^2x) sinx dx
= -2u/(1+u^2) du
let v = 1+u^2, so dv = 2u du
= -dv/v
To evaluate the integral of (sin 2x)/(1+cos^2 x), we start by using the double-angle identity for sine: sin 2x = 2sin x cos x. Thus, we can rewrite the integral as:
∫ (2sin x cos x)/(1+cos^2 x) dx
Next, we use the substitution u = cos x, which implies du = -sin x dx. Rearranging this equation, we have dx = -du/sin x.
Substituting u = cos x and dx = -du/sin x into the integral, we get:
∫ (2 sin x cos x)/(1 + cos^2 x) dx = ∫ (2u)/(1 + u^2) * (-du/sin x)
Now, notice that sin x appears in the denominator. Since we have du = -sin x dx, we can substitute it in the integral:
∫ (2u)/(1 + u^2) * (-du/sin x) = -∫ (2u)/(1 + u^2) * (du/du)
Simplifying this expression, we obtain:
-∫ 2u/(1 + u^2) du
Integrating this expression with respect to u, we get:
-ln(1 + u^2) + C
Finally, substituting back u = cos x, we arrive at the result:
-ln(1 + cos^2 x) + C
So, the integral of (sin 2x)/(1+cos^2 x) is equal to -ln(1 + cos^2 x) + C.