Integration by Parts (IBP) comes from the product rule: d(uv) = u dv + v du, giving integral u dv = uv - integral v du. The challenge is choosing u and dv wisely.
LIATE Rule: Choose u from this priority: Logarithmic (ln x, log x), Inverse trig (arctan x, arcsin x), Algebraic (x^n, polynomials), Trigonometric (sin x, cos x), Exponential (e^x, a^x). The function with higher priority becomes u.
Tabular Method (Repeated By-Parts): For integral x^n * e^(ax) dx or integral x^n * sin(ax) dx, create two columns: derivatives of the algebraic part and integrals of the other part. Alternate signs (+, -, +, -...) and multiply diagonally until the algebraic part reaches zero.
Example: integral x^3 * e^x dx
| Derivatives | Integrals | Sign |
|---|---|---|
| x^3 | e^x | + |
| 3x^2 | e^x | - |
| 6x | e^x | + |
| 6 | e^x | - |
| 0 | e^x |
Result: e^x(x^3 - 3x^2 + 6x - 6) + C
Cyclic By-Parts: For integral e^(ax)*sin(bx) dx or integral e^(ax)*cos(bx) dx, applying IBP twice brings back the original integral. Let I denote the original integral, then solve algebraically:
- I = e^(ax)(asin(bx) - b*cos(bx))/(a^2+b^2) + C
- integral e^(ax)cos(bx) dx = e^(ax)(acos(bx) + bsin(bx))/(a^2+b^2) + C
The e^x Shortcut: integral e^x[f(x) + f'(x)] dx = e^xf(x) + C. This is the most tested IBP-related result in JEE. To apply: (1) Check if the non-exponential part can be split as g(x) + g'(x), (2) If yes, the answer is e^xg(x) + C.