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Find the following integral: ∫ x sin(x) dx
This question is a good candidate for the integration by parts method, as it is the product of two different 'parts'.
Recall that if you have an integral of the form
∫ u(dv/dx) dx
it can be written as
uv – ∫ v(du/dx) dx.
We need to decide which part we will differentiate (as in, which part is u), and which part we will integrate (as in, which part is dv/dx).
We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1.
From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x) ). So, let
u = x, which implies du/dx = 1
dv/dx = sin(x). Integrating this to get v gives v = –cos(x).
So our integral is now of the form required for integration by parts.
∫ x sin(x) dx
= ∫ u(dv/dx) dx
= uv – ∫ v(du/dx) dx
= –x cos(x) – ∫ –cos(x)*1 dx
= –x cos(x) – ∫ –cos(x) dx
= –x cos(x) + ∫ cos(x) dx
The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x). Finally, as with all integration without limits, there must be a constant added, which I'll call c. So the final answer is
∫ x sin(x) dx = –x cos(x) + sin(x) + csee more