how to solve ∫ (sin^3/5 x) dx ?
To solve the integral of sin^(3/5)(x) dx, we can use a u-substitution technique.
First, let's rewrite the integral in a more simplified form. Since sin(x)^(3/5) can be expressed as (sin(x))^(3/5), the integral becomes:
∫ (sin(x))^(3/5) dx
Now, let's set u equal to sin(x).
u = sin(x)
Now, we need to find the derivative of u with respect to x.
du/dx = cos(x)
To solve for dx, we'll rearrange the equation:
dx = du / cos(x)
Substituting the values of u and dx in terms of x, we can rewrite the integral as:
∫ (sin(x))^(3/5) dx = ∫ u^(3/5) (du/cos(x))
Notice that cos(x) is present in the denominator. To eliminate it, we can use the identity cos(x) = √(1 - sin^2(x)).
Now, rewrite the denominator as:
∫ u^(3/5) (du/√(1 - sin^2(x)))
Next, we'll need to replace sin^2(x) with (1 - cos^2(x)) according to the above identity.
∫ u^(3/5) (du/√(1 - (1 - u^2)))
Simplifying further:
∫ u^(3/5) (du/√u^2)))
This can be simplified more:
∫ u^(3/5) / u du
We can combine the terms and rewrite the integral as:
∫ u^(-2/5) du
To integrate u^(-2/5), we'll use the power rule. The power rule states that to integrate x^n with respect to x, when n is not equal to -1, we can use the formula (x^(n+1))/(n+1).
For our integral:
∫ u^(-2/5) du = (u^(-2/5 + 1))/(-2/5 + 1)
Simplifying:
= (u^(3/5))/(3/5)
= 5/3 * u^(3/5)
Now, substitute u back in terms of x:
= 5/3 * sin(x)^(3/5)
Therefore, the solution to the integral ∫ (sin^(3/5)(x)) dx is:
5/3 * sin(x)^(3/5) + C
where C represents the constant of integration.