integral of 2sin^3xcosx dx
that is a sin cubed x. i think i use u-substituion, but im not quite sure how to do it. please help and thank you!
Whenever there is an odd power of sin or cos in a product, you could use the substition u=sin(x) or u=cos(x).
This is a case of u=sin(x), since
du=cos(x)dx
Substituting in the given expression gives
∫u³du
If you are working on a definite integral, do not forget to substitute the limits as well.
I'm sure you can take it from here.
To integrate the function ∫2sin^3(x)cos(x) dx using u-substitution, you can follow these steps:
Step 1: Identify the u-substitution variable. In this case, let u = sin(x).
Step 2: Calculate the derivative of u with respect to x. du/dx = cos(x).
Step 3: Solve for dx in terms of du. Rearrange the equation from Step 2 to get dx = du/cos(x).
Step 4: Rewrite the original integral in terms of u and du. Substitute sin(x) with u and dx with du/cos(x):
∫2sin^3(x)cos(x) dx = ∫2u^3 * (du/cos(x)).
Step 5: Simplify the expression. Substitute cos(x) with √(1 - sin^2(x)).
∫2u^3 * (du/cos(x)) = ∫2u^3 * (du/√(1 - u^2)).
Step 6: Integrate the expression with respect to u. Using the power rule of integration, we have:
∫2u^3 * (du/√(1 - u^2)) = 2 * ∫u^3/√(1 - u^2) du.
Step 7: Use a different u-substitution to simplify the integral further. Let v = 1 - u^2.
Step 8: Calculate the derivative of v with respect to u. dv/du = -2u.
Step 9: Solve for du in terms of dv. Rearrange the equation from Step 8 to get du = -(1/2) * dv/u.
Step 10: Rewrite the integral in terms of v and dv. Substitute 1 - u^2 with v and du with -(1/2) * dv/u:
2 * ∫u^3/√(1 - u^2) du = -2 * ∫u^3/(√(1 - u^2)) * (1/2) * dv/u.
Step 11: Simplify the expression. Cancel out the u's and rewrite the integral:
-2 * ∫(1/2) * v^(-1/2) dv.
Step 12: Integrate the expression with respect to v. Using the power rule of integration, we have:
-2 * ∫(1/2) * v^(-1/2) dv = -v^(1/2).
Step 13: Substitute back the original variable. Replace v with 1 - u^2:
-(-v^(1/2)) = v^(1/2) = (1 - u^2)^(1/2).
Therefore, the final solution is:
∫2sin^3(x)cos(x) dx = (1 - u^2)^(1/2) + C.
Remember to include the constant of integration (C) when finding the antiderivative.