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Wize AP Calculus (BC) Textbook > Integration Techniques

Integration by Parts

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Integration by Parts (IBP)

There is no "Product Rule for integrals" like there is for derivatives. When trying to integrate the product of two functions, we can use the technique of Integration By Parts.

∫u⋅dv=u⋅v−∫v⋅du\boxed{\displaystyle \int u \cdot dv=u \cdot v-\int v \cdot du }

When to use IBP

We can use IBP when integrating a product of two different function types.
  • ∫01xe5x dx\int_0^1xe^{5x}\ dx
  • ∫x2sin⁡x dx\int_{ }^{ }x^2\sin x\ dx
  • ∫e3xcos⁡(5x)dx\int_{ }^{ }e^{3x}\cos\left(5x\right)dx
  • ∫ln⁡x dx\int_{ }^{ }\ln x\ dx


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Procedure for IBP

  1. Identify the two different function types
  2. Let one part be uuand the other be dvdv.
  • L - Logarithmic Functions
  • I - Inverse Functions
  • A- Algebraic Functions
  • T- Trig Functions
  • E - Exponential Functions
  1. Differentiate u→duu \rightarrow duand integrate dv→vdv \rightarrow v
  2. Rewrite the integral using the formula ∫u⋅dv=u⋅v−∫v⋅du\displaystyle \int u \cdot dv=u \cdot v-\int v \cdot du
Watch Out!
Sometimes you may need to use IBP more than once in the same problem!


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Example: (Repeated) Integration By Parts

Compute the following indefinite integral

∫x2sin⁡x  ⁣dx{\displaystyle\int}x^2\sin x\de{x}

1. Identify the 2 function types
There is an algebra x2x^2 and a trig sin⁡x\sin x.

2. Pick u and dv
Let u=x2u=x^2 and dv=sin⁡x dxdv=\sin x\ dx.

3. Differentiate and integrate
  ⁣du=2x  ⁣dx\de{u}=2x\de{x} and v=−cos⁡(x)v=-\cos(x)

4. Rewriting using the formula:
∫x2sin⁡(x)  ⁣dx{\displaystyle\int}x^2\sin (x)\de{x}
=x2(−cos⁡x)−∫−cos⁡x(2x)dx=x^2(-\cos x)-{\displaystyle \int}-\cos x(2x)dx
=−x2cos⁡x+2∫x⋅cos⁡x dx=-x^2\cos x+2{\displaystyle \int}x\cdot\cos x\ dx

The integral looks a little simpler than before, but we still have a product of two different function types. So we need to use I by P again:
1. algebra: xx and trig: cos⁡x\cos x
2. Let u=xu=x and dv=cos⁡x dxdv=\cos x\ dx
3. Then du=dxdu=dx and v=sin⁡xv=\sin x
4. The new integral becomes

=−x2cos⁡x+2[xsin⁡x−∫sin⁡x dx]=-x^2\cos x+2\left[x\sin x-\int_{ }^{ }\sin x\ dx\right]
=−x2cos⁡x+2xsin⁡x+2cos⁡x+C=-x^2\cos x+2x\sin x+2\cos x+C
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Example: Integration By Parts

Evaluate the following indefinite integral

∫ln⁡(x2) dx \displaystyle \int_{ }^{ }\ln\left(\frac{x}{2}\right)\ dx\

1. ln/log: ln⁡(x2)\ln\left(\frac{x}{2}\right) and algebra: 11 (*tricky)

2. Let u=ln⁡(x2)u=\ln\left(\frac{x}{2}\right) and dv=1 dxdv=1\ dx

3. Then

du=1x2(12)dx=1xdx\displaystyle du=\frac{1}{\frac{x}{2}}\left(\frac{1}{2}\right)dx=\frac{1}{x}dx and v=xv=x

4. The new integral becomes:

=ln⁡(x2)x−∫x(1x)dx\displaystyle =\ln\left(\frac{x}{2}\right)x-\int_{ }^{ }x\left(\frac{1}{x}\right)dx
=xln⁡(x2)−∫1 dx\displaystyle =x\ln\left(\frac{x}{2}\right)-\int_{ }^{ }1\ dx
=xln⁡(x2)−x+C\displaystyle =x\ln\left(\frac{x}{2}\right)-x+C
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Example: (Repeated) Integration By Parts

Evaluate the following indefinite integral

∫e3xcos⁡5x   ⁣dx{\displaystyle\int} e^{3x}\cos5x\ \de{x}

1. Exponential: e3xe^{3x} and Trig: cos⁡5x\cos5x

2. Let u=cos⁡5xu=\cos5x and dv=e3x dxdv=e^{3x}\ dx

3. Then du=−5sin⁡5x dxdu=-5\sin5x\ dx and v=e3x3v=\frac{e^{3x}}{3}

4. The new integral becomes:
=cos⁡5x e3x3−∫(e3x3)(−5sin⁡5x dx) \displaystyle =\frac{\cos5x\ e^{3x}}{3}-\int_{ }^{ }\left(\frac{e^{3x}}{3}\right)\left(-5\sin5x\ dx\right)

=e3xcos⁡5x3+53∫e3xsin⁡5x dx\displaystyle =\frac{e^{3x}\cos5x}{3}+\frac{5}{3}\int_{ }^{ }e^{3x}\sin5x\ dx

Notice that the integral has a product of two different function types, so we use I by P again:
1. Exponential: e3xe^{3x} and Trig: sin⁡5x\sin5x

2. Let u=sin⁡5xu=\sin5x and dv=e3x dxdv=e^{3x}\ dx

3. Then du=5cos⁡5x dxdu=5\cos5x\ dx and v=e3x3v=\frac{e^{3x}}{3}

4. The new integral becomes:
=e3xcos⁡5x3+53[sin⁡5x(e3x3)−∫(e3x3)(5cos⁡5x dx)]\displaystyle =\frac{e^{3x}\cos5x}{3}+\frac{5}{3}\left[\sin5x\left(\frac{e^{3x}}{3}\right)-\int_{ }^{ }\left(\frac{e^{3x}}{3}\right)\left(5\cos5x\ dx\right)\right]

=e3xcos⁡5x3+59e3xsin⁡5x−259∫e3xcos⁡5x dx\displaystyle =\frac{e^{3x}\cos5x}{3}+\frac{5}{9}e^{3x}\sin5x^{ }-\frac{25}{9}\int_{ }^{ }e^{3x}\cos5x\ dx

The new integral looks like the original question.
Notice that this seems to repeat over and over again.
The trick here is to let I=∫e3xcos⁡5x dxI={\displaystyle \int}e^{3x}\cos5x\ dx (original integral).
We can rewrite:
I=e3xcos⁡5x3+59e3xsin⁡5x−259I\displaystyle I=\frac{e^{3x}\cos5x}{3}+\frac{5}{9}e^{3x}\sin5x^{ }-\frac{25}{9}I

I+259I=e3xcos⁡5x3+59e3xsin⁡5x\displaystyle I+\frac{25}{9}I=\frac{e^{3x}\cos5x}{3}+\frac{5}{9}e^{3x}\sin5x

349I=e3xcos⁡5x3+59e3xsin⁡5x\displaystyle\frac{34}{9}I=\frac{e^{3x}\cos5x}{3}+\frac{5}{9}e^{3x}\sin5x

I=934(e3xcos⁡5x3+59e3xsin⁡5x)\displaystyle I=\frac{9}{34}\left(\frac{e^{3x}\cos5x}{3}+\frac{5}{9}e^{3x}\sin5x\right)

or I=3e3xcos⁡5x+5e3xsin⁡5x34\displaystyle I=\frac{3e^{3x}\cos5x+5e^{3x}\sin5x}{34}

Therefore, ∫e3xcos⁡5x dx=3e3xcos⁡5x+5e3xsin⁡5x34+C\displaystyle \int e^{3x}\cos5x\ dx=\frac{3e^{3x}\cos5x+5e^{3x}\sin5x}{34}+C

Practice: Integration By Parts

Evaluate the following indefinite integral

∫xarctan⁡x  ⁣dx{\displaystyle \int}x\arctan x\de{x}

Practice: Integration By Parts

Evaluate the following indefinite integral

∫sec⁡3x   ⁣dx{\displaystyle\int}\sec^3x \space\de{x}