The u-substitution method replaces part of our integral with a variable so that it is easier to integrate. This can apply if we have large exponents or rational expressions.
Consider \(\int (2t-3)^{29} dt \). Let us use the u-substitution method to replace \( 2t-3 \) with \( u \).
\( \Longrightarrow \) differentiating both sides, we obtain \( 2 = \frac{du}{dt} \rightarrow dt = \frac{du}{2} \)
\( \Longrightarrow \) substituting for \( 2t-3 \) and \( dt \): \[ \int \frac{u^{29}}{2} du \]
\[ \Longrightarrow \ = \frac{1}{2}\frac{u^{30}}{30} + C \]
\[\ \ \ \ \ \ \ \ \ \ \Longrightarrow \ = \frac{1}{60}(2t-3)^{30} + C \]
Integration by parts is a method that used for integrating products of two functions, represented by the following formula: \[ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx \]
When approaching an integral with this method, the first step is determining which part of our integral is easiest to integrate by itself so that from our \( g'(x) \) we can figure out our \( g(x) \) in the formula.
The general rule when choosing \( f(x) \) is to follow this order:
I nverse Trigonometric
L ogs
A lgebraic
T rigonometric
E xponential
Before continuing onto our example, recall the integrals for our trigonometric functions:
\[ \int tanxdx = -ln(cosx) + C \ \ \ \ \ \ \ \ \ \ \int cotx dx = ln(sinx) + C \]
\[ \int secx dx = ln(secx + tanx) + C \ \ \ \ \ \ \ \ \ \ \int cscx dx = -ln(cscx + cotx) + C \]
Solve the following indefinite integral: \( \int xsin(x) dx \).
\( \Longrightarrow \) As per our ILATE structure, we choose \( f(x) = x \) and \( g '(x) = sin(x) \)
\[ \Longrightarrow \ \int xsin(x) dx = x(-cos(x)) - \int (1)(-cos(x)) dx \]
\[ \Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ = -xcos(x) - (-sin(x)) + C \ \ \ \ \ \ \ \ \ \]
\[ \Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -xcos(x) + sin(x) + C \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]
EXAMPLE
1. Solve for the indefinite integral \( \int_{a}^{b} e^x sin(x) dx \)
2. Solve for the definite integral \( \int_{0}^{1} x e^x dx \)
See the video below for the solution: