sorry but, this is my question :
how to solve ∫(sin ^ -3/2 x) dx ?
To solve the integral ∫(sin^(-3/2)(x)) dx, we can use a combination of the substitution method and trigonometric identities.
Step 1: Simplify the integrand
The first step is to simplify the integrand by using a trigonometric identity. Recall the identity: sin^2(x) + cos^2(x) = 1. Rearranging this identity, we get sin^2(x) = 1 - cos^2(x). Now, let's rewrite sin^(-3/2)(x) in terms of cos(x).
sin^(-3/2)(x) = (1/sin^3/2(x))
= (1/sqrt(sin^2(x)))
= (1/sqrt(1 - cos^2(x)))
Step 2: Make a substitution
Next, we can make a substitution to simplify the integral further. Let's substitute u = cos(x).
Differentiating both sides with respect to x, we get du = -sin(x) dx.
Thus, dx = -du/sin(x).
Using this substitution, we can rewrite the integral:
∫(sin^(-3/2)(x)) dx = -∫(1/sqrt(1 - cos^2(x))) * (du/sin(x))
= -∫(1/sqrt(1 - u^2)) * (du/sqrt(1 - u^2))
= -∫(1/(1 - u^2)) du
Step 3: Evaluate the integral
The new integral, -∫(1/(1 - u^2)) du, is now in a form that can be solved more easily. This integral is a standard integral called the integral of a rational function. It can be solved using partial fractions or by recognizing that it's the integral of the inverse tangent (arctan) function.
∫(1/(1 - u^2)) du = arctan(u) + C
Step 4: Substitute back for u
Finally, substitute back the value of u from our substitution (u = cos(x)) into the result:
arctan(u) + C = arctan(cos(x)) + C
So, the solution to the integral ∫(sin^(-3/2)(x)) dx is arctan(cos(x)) + C, where C is the constant of integration.