How do you integrate #int xsin2x# by integration by parts method?

Answer 1

#int xsin2xdx = -1/2xcos2x +1/4sin2x#

As:

#d(cos2x) =-2sin2x dx#

we can integrate by parts in this way:

#int xsin2xdx = -1/2 int x d(cos2x) = -1/2 xcos2x +1/2 int cos2x dx= -1/2xcos2x +1/4sin2x#
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Answer 2

To integrate ∫xsin(2x) dx using integration by parts method, follow these steps:

  1. Let u = x and dv = sin(2x) dx.
  2. Calculate du and v.
    • Differentiate u to get du.
    • Integrate dv to get v.
  3. Apply the integration by parts formula: ∫u dv = uv - ∫v du.
  4. Substitute the values of u, v, du, and dv into the formula.
  5. Evaluate the integral ∫v du.
  6. Combine the results to get the final answer.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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