Question

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.

Answer 1

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

Answer 2

∫ (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~ :)

Answer 3

*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