how to solve ∫(sin ^ -3/2 x) dx ?
To solve the integral ∫ (sin^(-3/2)(x)) dx, we need to apply the appropriate integration technique. In this case, we can use the substitution method.
Let's start by selecting an appropriate substitution. In this case, a good choice is to let u = sin(x), because then du/dx = cos(x) dx. We can rewrite the integral using this substitution:
∫ (sin^(-3/2)(x)) dx = ∫ (sin^(-3/2)(x)) * cos(x) dx.
Now, we'll substitute u = sin(x), so we have:
du/dx = cos(x) dx,
dx = du / cos(x).
Substituting this into the integral, we get:
∫ (u^(-3/2)) * (du / cos(x)).
Next, we simplify the expression:
∫ (u^(-3/2)) * (du / cos(x)) = ∫ (u^(-3/2)) * (1 / cos(x)) du.
Now we can focus on integrating with respect to u:
∫ (u^(-3/2)) * (1 / cos(x)) du = ∫ (1 / (u^(3/2) * cos(x))) du.
To solve this integral, we can use the power rule for integration. The integral of u^n with respect to u is given by (u^(n+1)) / (n+1), except when n = -1. In that case, the integral becomes the natural logarithm of u.
In our case, n = -3/2, so the integral becomes:
∫ (1 / (u^(3/2) * cos(x))) du = (u^(-1/2)) / (-1/2) = -2u^(-1/2) + C.
Now, we need to undo the substitution by replacing u with sin(x):
-2u^(-1/2) + C = -2(sin(x))^(-1/2) + C.
Therefore, the solution to the integral ∫ (sin^(-3/2)(x)) dx is -2(sin(x))^(-1/2) + C, where C is the constant of integration.
Please note that this is a general procedure, and it's always important to check the solution by taking the derivative of the obtained function to see if it matches the original function.