Hi, could you help me with this one please?
Evaluate
∫sin(x)cos^2(x)dx by using the substitution u=sinx
Ive done this by using u=cosx no problem but can't get this one :(
Thank you so much in advance!
by simple inspection, and knowing the rules of derivatives for sine and cosine ....
no need for substitution.
∫sin(x)cos^2(x)dx
= (-1/3) cos^3 x + c
Awkward choice, but let
u = sinx
du = cosx dx
∫sinx cos^2x dx
= ∫sinx cosx cosx dx
= ∫u√(1-u^2) du
Now let
v = 1-u^2
dv = -2u du
and your integral now becomes
∫-1/2 √u du
= (-1/2)(2/3)u^(3/2)
= -1/3 u^(3/2)
= -1/3 sin^3 x + C
Now you just have to show that the two answers are equivalent (differ only by using different values of C)
Thank you so much! ^^
My typo made it wrong. I hope you caught it. Let's pick up at
Now let
v = 1-u^2
dv = -2u du
and your integral now becomes
∫-1/2 √v dv
= (-1/2)(2/3)v^(3/2)
= -1/3 v^(3/2)
-1/3 (1-u^2)^(3/2)
= -1/3 cos^3 x + C
That's better!
Of course! I'll be happy to help you with this integration problem.
To evaluate ∫sin(x)cos^2(x)dx using the substitution u = sin(x), we can follow these steps:
Step 1: Start by differentiating the new equation u = sin(x) with respect to x to find du/dx.
du/dx = cos(x)
Step 2: Rearrange the equation to solve for dx:
dx = du/cos(x)
Step 3: Substitute the new variables and expression into the original integral:
∫sin(x)cos^2(x)dx = ∫u * (cos^2(x)) * (du/cos(x))
Step 4: Simplify the integral:
∫u * (cos^2(x)) * (du/cos(x)) = ∫u * cos(x) du
Step 5: Integrate the simplified integral:
∫u * cos(x) du = ∫u du
= (1/2)u^2 + C
Step 6: Substitute the original variable back in for u:
= (1/2)(sin(x))^2 + C
So, the final result is:
∫sin(x)cos^2(x)dx = (1/2)(sin(x))^2 + C
Therefore, the integral of sin(x)cos^2(x)dx, using the substitution u = sin(x), is equal to (1/2)(sin(x))^2 + C, where C is the constant of integration.