Evaluate ∫ (cos(x))^(1/2)sin(x)dx
Let u = cos(x)?
∫ (u)^(1/2)sin(x)dx = ∫ [2u^(3/2)/3]sin(x)dx
∫ [2cos(x)^(3/2)/3] (-cos(x)) dx?
I thought this involved the FTC, but now I'm thinking that's false.
ok, let u = cos x
then du = -sin x dx
then you have
∫ (u)^(1/2)(-du)
=-(2/3) u^(3/2) + c
of course that is
-(2/3) (cos x)^3/2 + c
Your problem seems to be not replacing dx with a calculated du
To evaluate the integral ∫ (cos(x))^(1/2)sin(x)dx, you correctly made the substitution u = cos(x).
Let's go through the steps again:
1. Let u = cos(x). Then, differentiate both sides with respect to x to get du = -sin(x)dx.
2. Substitute u = cos(x) and -sin(x)dx = du into the integral:
∫ (cos(x))^(1/2)sin(x)dx = ∫ (u)^(1/2)(-du) = -∫ u^(1/2) du.
3. Now, we can integrate ∫ u^(1/2) du. Recall the power rule for integrals: ∫ x^n dx = (x^(n+1))/(n+1) + C. Applying this rule, we get:
-∫ u^(1/2) du = -(2/3)u^(3/2) + C.
4. Finally, substitute u = cos(x) back into the result:
-(2/3)u^(3/2) + C = -(2/3)(cos(x))^(3/2) + C.
Therefore, the integral evaluates to -(2/3)(cos(x))^(3/2) + C, where C represents the constant of integration.
So, you were correct in making the u-substitution, and the integral does not involve the Fundamental Theorem of Calculus (FTC) in this case.