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CIE A-Level Maths Study Notes

2.5.5 Integration by Parts

Integration by parts is a powerful technique for integrating products of functions, which is based upon the product rule for differentiation. It allows us to transform the integral of a product into simpler integrals that we can evaluate more easily.

Fundamental Formula

Use for differentiable functions uu and vv:

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

For definite integrals from aa to bb:

abudv=[uv]ababvdu\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du

Choose uu using LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential).

Example Problems

Example 1: Integrating xsin(2x)x \sin(2x)

1. Identify uu and dvdv:

  • u=xu = x, dv=sin(2x)dxdv = \sin(2x) dx.

2. Differentiate and Integrate:

  • du=dxdu = dx, v=sin(2x)dx=12cos(2x)v = \int \sin(2x) dx = -\frac{1}{2} \cos(2x).

3. Apply the Formula:

  • xsin(2x)dx=x(12cos(2x))12cos(2x)dx\int x \sin(2x) dx = x \left(-\frac{1}{2} \cos(2x)\right) - \int -\frac{1}{2} \cos(2x) dx.

4. Simplify:

  • Result: 12xcos(2x)+14sin(2x)+C-\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C.

Example 2: Integrating x2exx^2 e^{-x}

1. Identify uu and dvdv:

  • u=x2u = x^2, dv=exdxdv = e^{-x} dx.

2. Differentiate and Integrate:

  • du=2xdxdu = 2x dx, v=exdx=exv = \int e^{-x} dx = -e^{-x}.

3. Apply the Formula:

  • x2exdx=x2ex2xexdx\int x^2 e^{-x} dx = -x^2 e^{-x} - \int -2x e^{-x} dx.

4. Simplify:

  • Result: (22xx2)ex+C(-2 - 2x - x^2) e^{-x} + C.

Example 3: Integrating ln(x)\ln(x)

1. Identify uu and dvdv:

  • u=ln(x)u = \ln(x), dv=dxdv = dx.

2. Differentiate and Integrate:

  • du=1xdxdu = \frac{1}{x} dx, v=xv = x.

3. Apply the Formula:

  • ln(x)dx=xln(x)x1xdx\int \ln(x) dx = x \ln(x) - \int x \cdot \frac{1}{x} dx.

4. Simplify:

  • Result: xln(x)x+Cx \ln(x) - x + C.

Example 4: Integrating xtan1(x)x \tan^{-1}(x)

1. Identify uu and dvdv:

  • u=tan1(x)u = \tan^{-1}(x), dv=xdxdv = x dx.

2. Differentiate and Integrate:

  • du=11+x2dxdu = \frac{1}{1 + x^2} dx, v=x22v = \frac{x^2}{2}.

3. Apply the Formula:

  • xtan1(x)dx=x22tan1(x)x2211+x2dx.\int x \tan^{-1}(x) dx = \frac{x^2}{2} \tan^{-1}(x) - \int \frac{x^2}{2} \cdot \frac{1}{1 + x^2} dx.

4. Simplify:

  • Result: x+(1+x2)tan1(x)2+C\frac{-x + (1 + x^2) \tan^{-1}(x)}{2} + C.

Worked Example:

For 14lnxxdx\int_1^4 \frac{\ln x}{\sqrt{x}} \,dx:

Solution:

1. Rewrite as 14lnxx1/2dx\int_1^4 \ln x \cdot x^{-1/2} \,dx.

2. Choose u=lnxu = \ln x, dv=x1/2dxdv = x^{-1/2}dx.

3. Compute du=1xdxdu = \frac{1}{x}dx, v=2x1/2v = 2x^{1/2}.

4. Apply formula: [2x1/2lnx]14142x1/2dx[2x^{1/2}\ln x]_1^4 - \int_1^4 2 \cdot x^{-1/2} \,dx.

5. Evaluate to get 4ln444\ln 4 - 4.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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