integrate
-sinx/�ã4+cos^2x dx
integrate
-sinx/�square root 4+cos^2x dx
I assume you mean
sin x/√(4+cos^2 x)
let u = cos x
du = -sin x dx
and you have
-∫1/√(4+u^2) du
standard hyperbolic trig substitution is used to get
-arcsinh(u/2)
= -arcsinh(cosx / 2)
To integrate the given expression ∫(-sinx)/(4+cos^2x) dx, we will use a substitution technique. Let's follow these steps:
Step 1: Substitute u = cos(x).
This substitution will simplify the expression and make it easier to integrate. Now, we need to find dx in terms of du.
Differentiate both sides of the substitution equation:
du/dx = -sin(x)
Rearrange the equation to solve for dx:
dx = du/(-sin(x))
Step 2: Substitute the values in the integral.
Now, replace sin(x) and dx in the integral with the corresponding values in terms of u.
∫(-sinx)/(4+cos^2x) dx = ∫(-du)/(4+u^2)
Step 3: Integrate the simplified expression.
The integral of (-du) / (4 + u^2) can be evaluated using the arctangent function.
∫(-du)/(4+u^2) = -arctan(u/2) + C
Step 4: Substitute back the original variable.
Substituting u = cos(x) back into the equation:
⇒ -arctan(u/2) + C
⇒ -arctan(cos(x)/2) + C
Therefore, the final solution is:
∫(-sinx)/(4+cos^2x) dx = -arctan(cos(x)/2) + C