Integral of 63x(cos(x))^2 dx

I got:
63x^2/4 +63sin(2x)/4 +63cos(2x)/8 +C

I keep getting this answer but the online site says it's wrong.

Wolfram agrees with you, they took out a common factor of 63

http://integrals.wolfram.com/index.jsp?expr=63x%28cos%28x%29%29%5E2&random=false

∫ (63x(cos(x))^2) dx

63 ∫ x(cos(x))^2 dx
Recall that cos^2 x is also equal to (1/2)(1+cos(2x)):
63/2 ∫ x(1+cos(2x)) dx
63/2 ∫ x + x cos(2x) dx
First term is easy, it becomes:
63/2 ∫ x dx = (63/4)x^2
For the second term, we use integration by parts (∫u dv = uv - ∫v du). Note the order of which term to substitute for u: LIATE = Logarithmic - Inverse Trigo - Algebraic - Trigonometric - Exponential. Thus, for 63/2∫ x cos(2x) dx
we let
u = x (this is algebraic)
du = dx
dv = cos(2x)dx
v = 1/2 sin(2x)
Second term becomes:
63/2 (x (1/2)(sin(2x)) - ∫ (1/2)sin(2x) dx)
(63/4)(x)*sin(2x) - (-63/4)*(1/2)cos(2x)
Adding all terms,
(63/4)x^2 + (63/4)(x)*sin(2x) + (63/8)*cos(2x)

Hope this helps~ :)

*sorry I forgot to add + C on the answer. ^^;

Anyway, in your answer that you showed us, there's no x multiplier in the second term. I think that's the only wrong term and the other terms are correct. :3

To find the integral of 63x(cos(x))^2 dx, you need to apply the standard techniques of integration, such as integration by parts or trigonometric substitution. Let's solve it using integration by parts.

First, let's identify the parts of the integrand to be used in the integration by parts formula: u and dv.
u = 63x
dv = (cos(x))^2 dx

To differentiate u, we use the power rule, which states that d/dx (x^n) = nx^(n-1).
du = d/dx(63x) = 63

To integrate dv, we have to simplify the integrand as much as possible. Here, we can use the trigonometric identity cos^2(x) = (1 + cos(2x))/2.
So, (cos(x))^2 = (1 + cos(2x))/2.

Now, let's integrate dv with this simplification:
∫ (cos(x))^2 dx = ∫ [(1 + cos(2x))/2] dx
= ∫ (1/2 + cos(2x)/2) dx
= ∫ (1/2) dx + ∫ (cos(2x)/2) dx
= (1/2)x + (1/4)sin(2x) + C1

Next, we can apply the integration by parts formula: ∫ u dv = uv - ∫ v du.
∫ 63x(cos(x))^2 dx = 63 * [(1/2)x + (1/4)sin(2x)] - ∫ (1/2) * 63 dx
= (63/2)x + (63/4)sin(2x) - (63/2)x + C2
= (63/4)sin(2x) + C.

So, the correct answer is: (63/4)sin(2x) + C.

It seems likely that there might have been an error in calculation or a mistake in the simplification of the integrand when you obtained your initial result.