Efficient Integration by Parts: Solving \int x^2*cosx dx with Step-by-Step Guide

Also, in your final answer, it should be (-x^2+2x+2) rather than (-x^2+2x-2).In summary, the integral of x^2*cos(x) dx can be solved using integration by parts by setting u=x^2 and dv=cos(x)dx. The final answer should be sin(x)(-x^2+2x+2) after factoring, but there was a mistake in the first step and the final answer should be (-x^2+2x+2).
  • #1
Aerosion
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Homework Statement



[tex]\int x^2*cosx dx[/tex]

Homework Equations


The Attempt at a Solution



Okay, so I started by making...
u=x^2
du=2x
dv=cos(x)
v=-sin(x)

Then I made the rudimentary equation:

[tex]\int x^2 * cos(x) dx = -x^2*sin(x) + \int 2x * sin(x) dx[/tex]

Then I took the last integration problem (the one all the way on the right) and did integration by parts on that one again, to make:

u=2x
du=2
dv=sin(x)
v=-cos(x)

to make...

[tex]\int x^2 * cos(x) dx = -x^2 * sin(x) + 2x * sin(x) + \int 2 * cos(x) dx[/tex]

I evaluated the last integration problem to be -2*sin(x), and got a final answer of sin(x)(-x^2+2x-2) (after factoring). Of course, my calculator says this is wrong, so where'd I mess up?

Thanks. (I'm getting better with Latex, btw)
 
Last edited:
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  • #2
In your first step, if dv=cos(x)dx, then v=sin(x) (no minus sign).
 

FAQ: Efficient Integration by Parts: Solving \int x^2*cosx dx with Step-by-Step Guide

What is integration by parts?

Integration by parts is a technique used in calculus to solve integrals that are products of two functions. It involves splitting the integrand into two parts and using a specific formula to integrate one part while differentiating the other.

How do I know when to use integration by parts?

Integration by parts is typically used when the integrand contains a product of two functions, one of which can be easily integrated while the other can be easily differentiated. It can also be used to simplify integrals that involve powers of trigonometric functions.

What is the formula for integration by parts?

The formula for integration by parts is: ∫ u dv = uv - ∫ v du, where u and v are the two parts of the integrand, and du and dv are their respective differentials.

How do I apply integration by parts to solve an integral?

To apply integration by parts, follow these steps:

  • Identify u and dv in the integrand.
  • Integrate u and differentiate dv.
  • Plug these values into the formula: ∫ u dv = uv - ∫ v du.
  • Solve for the integral on the right side of the equation.
  • Repeat the process until the integral is fully solved.

Can you provide an example of using integration by parts to solve an integral?

Yes, as an example, let's use integration by parts to solve ∫ x^2*cosx dx. First, we identify u = x^2 and dv = cosx dx. Then, we integrate u to get du = 2x dx and differentiate dv to get v = sinx. Plugging these values into the formula, we get ∫ x^2*cosx dx = x^2*sinx - ∫ 2x*sinx dx. We can continue to apply integration by parts to the second integral until we reach a point where we can easily solve the integral. Finally, we combine all the terms to get the final solution.

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