Math Problem Statement

intregation by parts

∫x^2sin(x)dx

Solution

The integral in the image is:

x2sin(x)dx\int x^2 \sin(x) \, dx

To solve this, we will use integration by parts. The formula for integration by parts is:

udv=uvvdu\int u \, dv = uv - \int v \, du

Step 1: Choose uu and dvdv

We let:

  • u=x2u = x^2 (so that du=2xdxdu = 2x \, dx)
  • dv=sin(x)dxdv = \sin(x) \, dx (so that v=cos(x)v = -\cos(x))

Step 2: Apply the formula

Using the integration by parts formula:

x2sin(x)dx=uvvdu\int x^2 \sin(x) \, dx = u v - \int v \, du

Substitute the values of uu, vv, dudu, and dvdv:

=x2(cos(x))(cos(x))(2xdx)= x^2 (-\cos(x)) - \int (-\cos(x)) (2x \, dx) =x2cos(x)+2xcos(x)dx= -x^2 \cos(x) + 2 \int x \cos(x) \, dx

Step 3: Apply integration by parts again

For the remaining integral xcos(x)dx\int x \cos(x) \, dx, we apply integration by parts again. Let:

  • u=xu = x (so that du=dxdu = dx)
  • dv=cos(x)dxdv = \cos(x) \, dx (so that v=sin(x)v = \sin(x))

Now apply the formula:

xcos(x)dx=xsin(x)sin(x)dx\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx =xsin(x)+cos(x)= x \sin(x) + \cos(x)

Step 4: Combine the results

Now substitute back into the original equation:

x2sin(x)dx=x2cos(x)+2(xsin(x)+cos(x))+C\int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2(x \sin(x) + \cos(x)) + C

Thus, the solution to the integral is:

x2cos(x)+2xsin(x)+2cos(x)+C\boxed{-x^2 \cos(x) + 2x \sin(x) + 2\cos(x) + C}

Would you like more details on any step?

Here are five related questions:

  1. How would you apply integration by parts to exsin(x)dx\int e^x \sin(x) \, dx?
  2. What are some common functions that require integration by parts?
  3. Can you solve x3cos(x)dx\int x^3 \cos(x) \, dx using integration by parts?
  4. How would you derive the reduction formula for xnexdx\int x^n e^x \, dx?
  5. How does integration by parts work with definite integrals?

Tip: Integration by parts is especially useful when integrating products of algebraic and trigonometric (or exponential) functions. Always carefully choose uu and dvdv to simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts

Formulas

∫ u dv = uv - ∫ v du

Theorems

Integration by Parts Theorem

Suitable Grade Level

Grades 11-12 (Advanced Calculus)