Lecture Notes: Integration by Parts

Introduction

• The integration of the function (2x \sin x) with respect to (x) doesn't match a basic integration formula.
• Substitution method is not suitable here.

Integration by Parts

• Formula: ( \int u , dv = uv - \int v , du )
• Goal: Divide the integral into parts (u) and (dv).

Choosing (u) and (dv)

• Select (u) such that differentiation simplifies it.
• In this case:
  • If (u = 2x), then (du = 2 , dx).
  • If (u = \sin x), then (du = \cos x , dx), which is not simpler.
• Conclusion: Choose (u = 2x) and (dv = \sin x , dx).

Determine (v)

• Integrate to find (v):
  • (v = \int \sin x , dx = -\cos x)

Applying Integration by Parts

• Substitute into the integration by parts formula:
  • Original integral: (\int 2x \sin x , dx)
  • Result: (u \cdot v - \int v , du)

Calculation

• (u \cdot v = -2x \cos x)
• (\int v , du = \int -2 \cos x , dx)
  • Factor out (-2):
  • (-2 \int \cos x , dx = -2 \sin x)
• Final result:
  • (-2x \cos x + 2 \sin x + C)

Simplification

• Factor out 2:
  • (2(\sin x - x \cos x) + C)

Conclusion

• The antiderivative is obtained using integration by parts.
• Mention of using integration by parts twice in more complex examples.

Summary

• This method allows solving integrals that are difficult to approach directly.
• Focus on choosing (u) and (dv) wisely for simplification.

Final Note

• Integration by parts can be applied multiple times for more complex integrals.
• Hope the explanation was clear and helpful.