Integration by Parts Technique
Smaridasa Mathematics – For the next integration technique, we will discuss Integration by Parts, which directly involves both "differentiation" and "integration". We use the Integration by Parts technique when the "algebraic substitution integration technique" cannot directly solve the integral.
Integration by Parts Formula
The formula for Integration by Parts is:
\[ \int u \, dv = uv - \int v \, du \]
In this formula, the given integral must be separated into two parts: one part is the function \( u \) and the other part (the function containing \( dx \)) is \( dv \). Therefore, this method is often called integration by parts.
Strategy for Choosing \( u \) and \( dv \):
To make integration by parts easier, we choose \( u \) as a function whose derivative simplifies or goes to zero, and choose \( dv \) as a part that is easy to integrate.
Example 1: Basic Integration by Parts
1. \( \displaystyle \int x \sqrt{x + 2} \, dx \)
Solution:
Choose \( u = x \) (since its derivative becomes 1, simpler), and \( dv = \sqrt{x+2} \, dx \).
\[ u = x \quad \Rightarrow \quad du = dx \]
\[ dv = \sqrt{x+2} \, dx \quad \Rightarrow \quad v = \int (x+2)^{1/2} dx = \frac{2}{3} (x+2)^{3/2} \]
Apply the formula:
\[ \int x \sqrt{x+2} \, dx = uv - \int v \, du \]
\[ = x \cdot \frac{2}{3} (x+2)^{3/2} - \int \frac{2}{3} (x+2)^{3/2} \, dx \]
\[ = \frac{2}{3} x (x+2)^{3/2} - \frac{2}{3} \cdot \frac{2}{5} (x+2)^{5/2} + c \]
\[ = \frac{2}{3} x (x+2)^{3/2} - \frac{4}{15} (x+2)^{5/2} + c \]
Example 2: Repeated Integration by Parts
2. \( \displaystyle \int x^2 \cos 2x \, dx \)
Solution (Standard Method):
First application: Choose \( u = x^2 \), \( dv = \cos 2x \, dx \).
\[ du = 2x \, dx, \quad v = \frac{1}{2} \sin 2x \]
\[ \int x^2 \cos 2x \, dx = \frac{1}{2} x^2 \sin 2x - \int x \sin 2x \, dx \]
Second application for \( \int x \sin 2x \, dx \): Choose \( u = x \), \( dv = \sin 2x \, dx \).
\[ du = dx, \quad v = -\frac{1}{2} \cos 2x \]
\[ \int x \sin 2x \, dx = -\frac{1}{2} x \cos 2x + \frac{1}{2} \int \cos 2x \, dx = -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x \]
Combine results:
\[ \int x^2 \cos 2x \, dx = \frac{1}{2} x^2 \sin 2x - \left( -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x \right) + c \]
\[ = \frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x + c \]
\[ = \left( \frac{1}{2} x^2 - \frac{1}{4} \right) \sin 2x + \frac{1}{2} x \cos 2x + c \]
Tabular Method (Tanzalin Method)
For repeated integration by parts, the Tabular Method (sometimes called the Tanzalin method) organizes the work. One function is differentiated repeatedly until it becomes zero, and the other is integrated repeatedly. Then, multiply diagonally with alternating signs.
Example 3: Tabular Method
3. Solve Example 2 using the Tabular Method: \( \displaystyle \int x^2 \cos 2x \, dx \)
Tabular setup:
| Differentiate (u) | Integrate (dv) |
|---|---|
| \( + \, x^2 \) | \( \cos 2x \) |
| \( - \, 2x \) | \( \frac{1}{2} \sin 2x \) |
| \( + \, 2 \) | \( -\frac{1}{4} \cos 2x \) |
| \( - \, 0 \) | \( -\frac{1}{8} \sin 2x \) (not needed) |
Multiply diagonally with alternating signs and sum:
\[ \int x^2 \cos 2x \, dx = (x^2)\left( \frac{1}{2} \sin 2x \right) - (2x)\left( -\frac{1}{4} \cos 2x \right) + (2)\left( -\frac{1}{8} \sin 2x \right) + c \]
\[ = \frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x + c \]
Example 4: Tabular Method with Higher Power
4. \( \displaystyle \int 2x^3 \cos x \, dx \)
Tabular method:
| Differentiate | Integrate |
|---|---|
| \( + \, 2x^3 \) | \( \cos x \) |
| \( - \, 6x^2 \) | \( \sin x \) |
| \( + \, 12x \) | \( -\cos x \) |
| \( - \, 12 \) | \( -\sin x \) |
| \( + \, 0 \) | \( \cos x \) (not needed) |
Multiply and sum:
\[ \int 2x^3 \cos x \, dx = (2x^3)(\sin x) - (6x^2)(-\cos x) + (12x)(-\sin x) - (12)(-\cos x) + c \]
\[ = 2x^3 \sin x + 6x^2 \cos x - 12x \sin x + 12 \cos x + c \]
\[ = (2x^3 - 12x) \sin x + (6x^2 + 12) \cos x + c \]
Example 5: Using Trigonometric Identities First
5a. \( \displaystyle \int 4x \sin x \cos x \, dx \)
Use identity: \( \sin 2x = 2 \sin x \cos x \), so \( 4x \sin x \cos x = 2x \sin 2x \).
Now integrate \( \int 2x \sin 2x \, dx \) by parts (tabular method):
| \( + \, 2x \) | \( \sin 2x \) |
| \( - \, 2 \) | \( -\frac{1}{2} \cos 2x \) |
| \( + \, 0 \) | \( -\frac{1}{4} \sin 2x \) (not needed) |
\[ \int 2x \sin 2x \, dx = (2x)\left( -\frac{1}{2} \cos 2x \right) - (2)\left( -\frac{1}{4} \sin 2x \right) + c \]
\[ = -x \cos 2x + \frac{1}{2} \sin 2x + c \]
5b. \( \displaystyle \int 2x \cos^2 x \, dx \)
Use identity: \( \cos^2 x = \frac{1}{2}(1 + \cos 2x) \).
\[ 2x \cos^2 x = x + x \cos 2x \]
\[ \int 2x \cos^2 x \, dx = \int x \, dx + \int x \cos 2x \, dx \]
First integral: \( \int x \, dx = \frac{1}{2} x^2 \).
Second integral \( \int x \cos 2x \, dx \) by tabular method:
| \( + \, x \) | \( \cos 2x \) |
| \( - \, 1 \) | \( \frac{1}{2} \sin 2x \) |
| \( + \, 0 \) | \( -\frac{1}{4} \cos 2x \) (not needed) |
\[ \int x \cos 2x \, dx = x \left( \frac{1}{2} \sin 2x \right) - (1)\left( -\frac{1}{4} \cos 2x \right) + c_1 \]
\[ = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + c_1 \]
Combine:
\[ \int 2x \cos^2 x \, dx = \frac{1}{2} x^2 + \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + c \]
5c. \( \displaystyle \int 6x \cos(3x) \cos(2x) \, dx \)
Use product-to-sum identity: \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \).
\[ 6x \cos(3x) \cos(2x) = 3x [\cos(5x) + \cos x] = 3x \cos 5x + 3x \cos x \]
Now integrate each separately by tabular method.
First: \( \int 3x \cos 5x \, dx \)
| \( + \, 3x \) | \( \cos 5x \) |
| \( - \, 3 \) | \( \frac{1}{5} \sin 5x \) |
| \( + \, 0 \) | \( -\frac{1}{25} \cos 5x \) (not needed) |
\[ = 3x \left( \frac{1}{5} \sin 5x \right) - 3 \left( -\frac{1}{25} \cos 5x \right) + c_1 = \frac{3}{5} x \sin 5x + \frac{3}{25} \cos 5x + c_1 \]
Second: \( \int 3x \cos x \, dx \)
| \( + \, 3x \) | \( \cos x \) |
| \( - \, 3 \) | \( \sin x \) |
| \( + \, 0 \) | \( -\cos x \) (not needed) |
\[ = 3x \sin x - 3 (-\cos x) + c_2 = 3x \sin x + 3 \cos x + c_2 \]
Combine:
\[ \int 6x \cos(3x) \cos(2x) \, dx = \frac{3}{5} x \sin 5x + \frac{3}{25} \cos 5x + 3x \sin x + 3 \cos x + c \]
Example 6: Combining Substitution and Parts
6. \( \displaystyle \int 2x^3 \cos(x^2) \, dx \)
First use substitution: let \( u = x^2 \), then \( du = 2x \, dx \).
\[ \int 2x^3 \cos(x^2) \, dx = \int x^2 \cos(x^2) \cdot 2x \, dx = \int u \cos u \, du \]
Now integrate \( \int u \cos u \, du \) by parts (tabular method):
| \( + \, u \) | \( \cos u \) |
| \( - \, 1 \) | \( \sin u \) |
| \( + \, 0 \) | \( -\cos u \) (not needed) |
\[ \int u \cos u \, du = u \sin u - ( - \cos u ) + c = u \sin u + \cos u + c \]
Substitute back \( u = x^2 \):
\[ = x^2 \sin(x^2) + \cos(x^2) + c \]