Evaluate ∫(2x^3/√(1-x^4))dx
I can't figure out what u and du is...
please help thanks
u= sqrt(1-x^4)
du= 1/2sqrt(1-x^4)*(-4x^3)dx
= -2x^3/sqrt(1-x^4) dx
so the integral is
INT -1/u du
or,
u = 1-x^4
du = -4x^3 du
so
∫(2x^3/√(1-x^4))dx
= -1/2 ∫1/√u du
= (-1/2)(2√u)
= -√(1-x^4)
Extra credit: Can you find the error in bobpursley's solution?
no, but thanks for your answer!
To evaluate the integral ∫(2x^3/√(1-x^4))dx, we can use the substitution method. Let's make the substitution u = 1 - x^4. Then, we can find du in terms of dx.
Differentiating both sides of u = 1 - x^4 with respect to x, we get:
du/dx = -4x^3.
Rearranging this equation, we have:
dx = - (1/4x^3) du.
Now, we have expressed dx in terms of du, which allows us to rewrite the integral in terms of u:
∫(2x^3/√(1-x^4))dx = ∫(2x^3/√u)dx.
Substituting dx = - (1/4x^3) du into the integral, we get:
∫(2x^3/√u) * (-1/4x^3) du = -1/8 ∫(1/√u) du.
Simplifying this further:
-1/8 ∫(1/√u) du = -1/8 * ∫u^(-1/2) du.
Now, we can use the power rule for integration, which states that:
∫x^n dx = (x^(n+1))/(n+1) + C,
where C is the constant of integration. Applying this rule, we have:
-1/8 * ∫u^(-1/2) du = -1/8 * (u^(1/2))/(1/2) + C.
Simplifying the expression, we get:
-1/8 * (u^(1/2))/(1/2) + C = -1/4 * √u + C.
Finally, substituting back u = 1 - x^4, we have:
∫(2x^3/√(1-x^4))dx = -1/4 * √(1 - x^4) + C.
Thus, the final result is -1/4 * √(1 - x^4) + C.