whats the indefinite integral of xcosx?

Question

whats the indefinite integral of xcosx

thanks!

thank you, the minute i put the question down i actually just figured it out for myself through that "U" formula. Thanks anyways :(

Answer 1

You need to do integration by parts, which is a little annoying.
The formula is:
∫ u dv = uv - ∫ v du

So in ∫ x cos(x) dx,
u = x
du = 1 dx = dx

dv = cos(x) dx
v = ∫ cos x dx = sin(x)

Now just plug into the formula:
∫ xcos(x) dx = xsin(x) - ∫ sin(x) dx

= xsin(x) - ( -cos(x) )

= cos(x) + xsin(x) + C

Answer 2

The C is a constant of integration. Think of it like this. (1/3)x^3 is a solution to the Integral (x^2 dx) because it is the antiderivative, but (1/3)x^3 + C is also a solution because the derivative of [(1/3)x^3 + C] = x^2, since the d/dx{C} = 0. So there is always a constant associated with the antiderivative.

Answer 3

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RE:
whats the indefinite integral of xcosx?
whats the indefinite integral of xcosx

thanks!

Answer 4

you need to use integration by parts. which is, if you are unfamiliar with it:

∫f(x)g'(x) = f(x)g(x) - ∫f'(x)g(x)

so let f(x) = x which means f'(x) = 1
and let g'(x) = cos(x), g(x) = sin(x)

hence,

∫xcos(x) = xsin(x) - ∫ 1(sin(x))
= xsin(x) + cos(x) + C

Answer 5

Using integration by parts:-
I = ∫ (u dv/dx) dx = u v - ∫ v.(du/dx).dx
Let u = x and dv/dx = cos x
du/dx = 1 , v = sin x
I = x.sinx - ∫ sin x.dx
I = x.sin x + cos x + C